Class 12 Chemistry NCERT Solutions (Chemical Kinetics) – Important Questions
Q 1. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(a)
\(\begin{array}{l}3\; NO(g) \rightarrow N_{2}O (g) \; Rate = k\left [ NO \right ]^{2}\end{array} \)
(b)
\(\begin{array}{l}H_{2} O_{2} (aq) + 3I^{-}(aq) + 2H^{+} \rightarrow 2H_{2}O (I) + I^{-} \; Rate = k \left [ H^{2}O^{2} \right ]\left [ I^{-} \right ]\end{array} \)
(c)
\(\begin{array}{l}CH_{3}CHO (g) \rightarrow CH_{4} (g) + CO (g) \; Rate = k\left [ CH_{3} CHO \right ]^{\frac{3}{2}}\end{array} \)
(d)
\(\begin{array}{l}C_{2}H_{5}Cl (g) \rightarrow C_{2}H_{4} (g) + HCl (g) \; Rate = k\left [ C_{2} H_{5}Cl \right ]\end{array} \)
Ans:
(a) Given rate =
\(\begin{array}{l}k\left [ NO \right ]^{2}\end{array} \)
Therefore, the order of the reaction = 2
Dimensions of
\(\begin{array}{l}k = \frac{Rate}{\left [ NO \right ]^{2}}\end{array} \)
\(\begin{array}{l}\\ = \frac{mol \; L^{-1} s^{-1}}{\left ( mol \; L^{-1} \right )^{2}} \\ \\ = \frac{mol \; L^{-1} s^{-1}}{mol^{2}\; L^{-2}} \\ \\ = L \; mol^{-1} s^{-1}\end{array} \)
(b) Given rate =
\(\begin{array}{l}k[ H_{2}O_{2} ][ I ^{-}]\end{array} \)
Therefore, the order of the reaction = 2
Dimensions of
\(\begin{array}{l}k = \frac{Rate}{\left [ H_{2}O_{2} \right ]\left [ I^{-} \right ]}\end{array} \)
\(\begin{array}{l}\\ = \frac{mol \; L^{-1} S^{-1}}{\left ( mol \; L^{-1} \right ) \left ( mol \; L^{-1} \right )} \\ \\ = L \; mol^{-1} s^{-1}\end{array} \)
(c) Given rate =
\(\begin{array}{l}= k \left [ CH_{3} CHO \right ]^{\frac{3}{2}}\end{array} \)
Therefore, the order of reaction =
\(\begin{array}{l}\frac{3}{2}\end{array} \)
Dimensions of
\(\begin{array}{l}k = \frac{Rate}{\left [ CH_{3} CHO \right ]^{\frac{3}{2}}} \\ \\ = \frac{mol \; L^{-1}s^{-1}}{\left (mol \; L^{-1} \right )^{\frac{3}{2}}} \\ \\ = \frac {mol\; L^{-1} s^{-1}}{mol^{\frac{3}{2}}\; L^{\frac{3}{2}}} \\ \\ L^{\frac{1}{2}}\; mol^{ -\frac{1}{2}} \; s^{-1}\end{array} \)
(d) Given rate =
\(\begin{array}{l}k = \left [ C_{2}H_{5}Cl \right ]\end{array} \)
Therefore, the order of the reaction = 1
Dimension of
\(\begin{array}{l}k = \frac{Rate}{\left [ C_{2}H_{5}Cl \right ]} \\ \\ = \frac{mol\; L^{-1} s^{-1}}{mol \; L^{-1}} \\ \\ = s^{-1}\end{array} \)
Q 2. For the reaction:
\(\begin{array}{l}2A + B \rightarrow A_{2}B\end{array} \)
is
\(\begin{array}{l}k\left [ A \right ]\left [ B \right ]^{2}\end{array} \)
with
\(\begin{array}{l}k = 2.0 \times 10^{-6}\; mol^{-2}L^{2} \; s^{-1}\end{array} \)
. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1
Ans:
The initial rate of reaction is
Rate =
\(\begin{array}{l}k\left [ A \right ]\left [ B \right ]^{2} \\ \\ = \left ( 2.0 \times 10^{-6} mol^{-2} L^{2} s^{-1} \right )\left ( 0.1 \; mol \; L^{-1} \right )\left ( 0.2 \; mol \; L^{-1} \right )^{2} \\ \\ = 8.0 \times 10^{-9} mol^{-2}L^{2} s^{-1}\end{array} \)
When [A] is reduced from
\(\begin{array}{l}0.1 \; mol\; L^{-1} \; to \; 0.06 \; mol\; L^{-1}\end{array} \)
, the concentration of A reacted =
\(\begin{array}{l}\left (0.1 – 0.06 \right ) \; mol\; L^{-1} = 0.04 \; mol\; L^{-1}\end{array} \)
Therefore, the concentration of B reacted
\(\begin{array}{l}= \frac {1}{2} \times 0.04 \; mol \; L^{-1} = 0.02 \; mol \; L^{-1}\end{array} \)
Then, the concentration of B available,
\(\begin{array}{l}\left [ B \right ] = \left ( 0.2 – 0.02 \right ) mol \; L^{-1} = 0.18\; mol \; L^{-1}\end{array} \)
After [A] is reduced to
\(\begin{array}{l}0.06 \; mol \; L^{-1}\end{array} \)
, the rate of the reaction is given by,
Rate =
\(\begin{array}{l}k \left [ A \right ]\left [ B \right ]^{2} \\ \\ = \left ( 2.0 \times 10^{-6} mol^{-2}L^{2}s^{-1} \right )\left ( 0.06\; mol L^{-1} \right )\left ( 0.18 \; mol \; L^{-1} \right )^{2} \\ \\ = 3.89\; \times 10^{-9} mol \; L^{-1} s^{-1}\end{array} \)
Q 3. The decomposition of NH3 on the platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s –1?
Ans:
The decomposition of NH3 on the platinum surface is represented by the following equation:
\(\begin{array}{l}2NH^{3(g)} \overset{Pt}{\rightarrow} N_{2(g)} + 3H_{2(g)}\end{array} \)
Therefore,
\(\begin{array}{l}Rate = -\frac{1}{2}\frac{d\left [ NH_{3} \right ]}{dt} = \frac{d\left [ N_{2} \right ]}{dt} = \frac{1}{3}\frac{d\left [ H_{2} \right ]}{dt}\end{array} \)
However, it is given that the reaction is of zero order.
Therefore,
\(\begin{array}{l}-\frac{1}{2}\frac{d\left [ NH_{3} \right ]}{dt} = \frac{d\left [ N_{2} \right ]}{dt} = \frac{1}{3}\frac{d\left [ H_{2} \right ]}{dt} = k \\ \\ = 2.5 \times 10^{-4}\; mol L^{-1} s^{-1}\end{array} \)
Therefore, the rate of production of
\(\begin{array}{l}N_{2}\end{array} \)
is
\(\begin{array}{l}\frac{d\left [N_{2} \right ]}{dt} = 2.5 \times 10^{-4} mol\;L^{-1}s^{-1}\end{array} \)
And, the rate of production of
\(\begin{array}{l}H_{2}\end{array} \)
is
\(\begin{array}{l}\frac{d\left [H_{2} \right ]}{dt} = 3 \times 2.5 \times 10^{-4} mol\;L^{-1}s^{-1} \\ \\ = 7.5 \times 10^{-4} \; mol \;L^{-1}s^{-1}\end{array} \)
Q 4. The decomposition of dimethyl ether leads to the formation of
\(\begin{array}{l}CH_{4}, H_{2}, \; and \; CO\end{array} \)
and the reaction rate is given by
\(\begin{array}{l}Rate = k\left [ CH_{3} O\, C\! H_{3} \right ]^{\frac{3}{2}}\end{array} \)
The rate of reaction is followed by an increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
\(\begin{array}{l}Rate = k\left ( P_{CH_{3} O\, C\! H_{3}} \right )^{\frac{3}{2}}\end{array} \)
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3] 3/2 The rate of reaction is followed by an increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, Rate = k p(CHOCH3)3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Ans:
If pressure is measured in bar and time in minutes, then
Unit of rate =
\(\begin{array}{l}bar \; min^{-1}\end{array} \)
\(\begin{array}{l}Rate = k\left ( P_{CH_{3} O\, C\! H_{3}} \right )^{\frac{3}{2}} \\ \\ \Rightarrow k = \frac{Rate}{k\left ( P_{CH_{3} O\, C\! H_{3}} \right )^{\frac{3}{2}}}\end{array} \)
Therefore, the unit of rate constants
\(\begin{array}{l}(k) = \frac{bar\; min^{-1}}{bar^{\frac{3}{2}}} \\ \\ = bar^{\frac{-1}{2}}min^{-1}\end{array} \)
Q 5. Mention the factors that affect the rate of a chemical reaction.
Ans:
The factors which are responsible for the effect on the chemical reaction’s rate are
(a) Reaction temperature
(b) Presence of a catalyst
(c) The concentration of reactants (pressure in case of gases)
(d) Nature of the products and reactants
(e) Radiation exposure
(f) Surface area
Q 6. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Ans:
Let the concentration of the reactant be [A] = a
Rate of reaction,
\(\begin{array}{l}R = k [A]^{2}\\ \\ = ka^{2}\end{array} \)
(a) If the concentration of the reactant is doubled, i.e., [A] = 2a, then the rate of the reaction would be
\(\begin{array}{l}R’ = k\left ( A \right )^{2} \\ \\ = 4 ka^{2} \\ \\ = 4 \; R\end{array} \)
Therefore, the rate of the reaction now will be 4 times the original rate.
(b) If the concentration of the reactant is reduced to half, i.e.,
\(\begin{array}{l}\left [ A \right ] = \frac{1}{2}a\end{array} \)
, then the rate of the reaction would be
\(\begin{array}{l}R” = k\left ( \frac{1}{2} a \right )^{2} \\ \\ = \frac{1}{4} k a \\ \\ = \frac{1}{4} R\end{array} \)
Therefore, the rate of the reaction will be reduced to
\(\begin{array}{l}\frac{1}{4} ^{th}\end{array} \)
Q 7. What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?
Ans:
When a temperature of
\(\begin{array}{l}10^{ \circ }\end{array} \)
rises for a chemical reaction, then the rate constant increases and becomes near to double its original value.
The temperature effect on the rate constant can be represented quantitatively by the Arrhenius equation.
\(\begin{array}{l}k = Ae^{ -E_{a} / RT}\end{array} \)
Where,
k = rate constant
A = Frequency factor / Arrhenius factor
R = gas constant
T = temperature
\(\begin{array}{l}E_{a}\end{array} \)
= activation energy for the reaction.
Q 8. In a pseudo-first-order reaction in water, the following results were obtained:
| t/s | 0 | 30 | 60 | 90 |
| [Ester]mol / L | 0.55 | 0.31 | 0.17 | 0.085 |
Calculate the average rate of reaction between the time interval of 30 to 60 seconds.
Ans:
(a) The avg rate of reaction between the time intervals, 30 to 60 seconds,
\(\begin{array}{l}= \frac{d\left [ Ester \right ]}{dt} \\ \\ = \frac{0.31 – 0.17}{60 – 30} \\ \\ = \frac{0.14}{30} \\ \\ = 4.67 \times 10^{-3}\; mol \; l^{-1}\; s^{-1}\end{array} \)
(b) For a pseudo-first-order reaction,
\(\begin{array}{l}\\k = \frac{2.303}{t} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]}\\ \\ For \; t = 30\; s \\ \\ k_{1} = \frac{2.303}{30} \log \frac{ 0.55}{ 0.31} \\ \\ = 1.911 \times 10^{-2} s^{-1} \\ \\ For \; t = 60\; s \\ \\ k_{2} = \frac{2.303}{60} \log \frac{ 0.55}{ 0.17} \\ \\ = 1.957 \times 10^{-2} s^{-1} \\ \\ For \; t = 90\; s \\ \\ k_{3} = \frac{2.303}{90} \log \frac{ 0.55}{ 0.085} \\ \\ = 2.075 \times 10^{-2} s^{-1}\end{array} \)
Then, the avg rate constant,
\(\begin{array}{l}k = \frac{k_{1} + k_{2} + k_{3}}{3} \\ \\ = \frac{\left (1.911 \times 10^{-2} \right ) + \left (1.957 \times 10^{-2} \right ) + \left (2.075 \times 10^{-2} \right )}{3} \\ \\ = 1.98 \times 10^{-2}\;s^{-1}\end{array} \)
Q 9. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected by increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Ans:
(a) The differential rate equation will be
\(\begin{array}{l}-\frac{d\left [ R \right ]}{dt} = k\left [ A \right ]\left [ B \right ]^{2}\end{array} \)
(b) If the concentration of B is increased three times, then
\(\begin{array}{l}-\frac{d\left [ R \right ]}{dt} = k\left [ A \right ]\left [ 3B \right ]^{2} \\ \\ = 9. k\left [ A \right ]\left [ B \right ]^{2}\end{array} \)
Therefore, the reaction rate will be increased by 9 times.
(c) When the concentrations of both A and B are doubled,
\(\begin{array}{l}-\frac{d\left [ R \right ]}{dt} = k\left [ 2A \right ]\left [ 2B \right ]^{2} \\ \\ = 8. k\left [ A \right ]\left [ B \right ]^{2}\end{array} \)
Therefore, the rate of reaction will increase 8 times.
Q10. In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
| \(\begin{array}{l}A/mol\; L^{-1}\end{array} \) | 0.20 | 0.20 | 0.40 |
| \(\begin{array}{l}B/mol\; L^{-1}\end{array} \) | 0.30 | 0.10 | 0.05 |
| \(\begin{array}{l}r_{0}/mol\; L^{-1}\; s^{-1}\end{array} \) | \(\begin{array}{l}5.07 \times 10^{-5}\end{array} \) | \(\begin{array}{l}5.07 \times 10^{-5}\end{array} \) | \(\begin{array}{l}1.43 \times 10^{-4}\end{array} \) |
What is the order of the reaction with respect to A and B?
Ans:
Let the order of the reaction with respect to A be x and with respect to B be y.
Then,
\(\begin{array}{l}r_{0} = k\left [ A \right ]^{x} \left [ B \right ]^{y} \\ \\ 5.07 \times 10^{-5} = k\left [ 0.20 \right ]^{x}\left [ 0.30 \right ]^{y} \;\;\;\; (i) \\ \\ 5.07 \times 10^{-5} = k\left [ 0.20 \right ]^{x}\left [ 0.10 \right ]^{y} \;\;\;\; (ii) \\ \\ 1.43 \times 10^{-4} = k\left [ 0.40 \right ]^{x}\left [ 0.05 \right ]^{y} \;\;\;\; (iii)\end{array} \)
Dividing equation (i) by (ii), we get
\(\begin{array}{l}\frac{5.07 \times 10^{-5} }{5.07 \times 10^{-5} } = \frac{k\left [ 0.20 \right ]^{x}\left [ 0.30 \right ]^{y} }{k\left [ 0.20 \right ]^{x}\left [ 0.10 \right ]^{y} } \\ \\ \Rightarrow 1 = \frac{\left [ 0.30 \right ]^{y}}{\left [ 0.10 \right ]^{y}} \\ \\ \Rightarrow \left ( \frac{0.30}{0.10} \right )^{0} = \left ( \frac{0.30}{0.10} \right )^{y} \\ \\ \Rightarrow y = 0\end{array} \)
Dividing equation (iii) by (ii), we get
\(\begin{array}{l}\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \frac{k\left [ 0.40 \right ]^{x}\left [ 0.05 \right ]^{y}}{k\left [ 0.20 \right ]^{x}\left [ 0.30 \right ]^{y}} \\ \\ \Rightarrow \frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \frac{\left [ 0.40 \right ]^{x}}{\left [ 0.20 \right ]^{x}} \;\;\;\;\; \begin{bmatrix} Since\; y = 0,\\ \left [ 0.05 \right ]^{y} = \left [ 0.30 \right ]^{y} = 1 \end{bmatrix} \\ \\ \Rightarrow 2.821 = 2^{x} \\ \\ \Rightarrow \log 2.821 = x \log 2 \;\;\;\;\; (taking\; log \; on \; both\; sides) \\ \\ \Rightarrow x = \frac{\log 2.821}{\log 2} \\ \\ = 1.496 \\ \\ = 1.5 \; (Approximately)\end{array} \)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.
Q 11. The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D
| Exp. | \(\begin{array}{l}\frac{A}{mol L^{-1}}\end{array} \) | \(\begin{array}{l}\frac{B}{mol L^{-1}}\end{array} \) | Initial rate of formation of \(\begin{array}{l}\frac{D}{mol\; L^{-1} \;min^{-1}}\end{array} \) |
| 1 | 0.1 | 0.1 | \(\begin{array}{l}6.0 \times 10^{-3}\end{array} \) |
| 2 | 0.3 | 0.2 | \(\begin{array}{l}7.2 \times 10^{-2}\end{array} \) |
| 3 | 0.3 | 0.4 | \(\begin{array}{l}2.88 \times 10^{-1}\end{array} \) |
| 4 | 0.4 | 0.1 | \(\begin{array}{l}2.4 \times 10^{-2}\end{array} \) |
Determine the rate law and the rate constant for the reaction.
Ans:
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, the rate of the reaction is given by,
Rate =
\(\begin{array}{l}k\left [ A \right ]^{x} \left [ B \right ]^{y}\end{array} \)
According to the question,
\(\begin{array}{l}6.0 \times 10^{-3} = k\left [ 0.1 \right ]^{x} \left [ 0.1 \right ]^{y}\end{array} \)
——- (1)
\(\begin{array}{l}7.2 \times 10^{-2} = k\left [ 0.3 \right ]^{x} \left [ 0.2 \right ]^{y}\end{array} \)
——–(2)
\(\begin{array}{l}2.88 \times 10^{-1} = k\left [ 0.3 \right ]^{x} \left [ 0.4 \right ]^{y}\end{array} \)
——–(3)
\(\begin{array}{l}2.4 \times 10^{-2} = k\left [ 0.4 \right ]^{x} \left [ 0.1 \right ]^{y}\end{array} \)
——–(4)
Dividing equation (4) by (1), we get
\(\begin{array}{l}\frac{2.4 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k\left [ 0.4 \right ]^{x} \left [ 0.1 \right ]^{y}}{k\left [ 0.1 \right ]^{x} \left [ 0.1 \right ]^{y}}\end{array} \)
\(\begin{array}{l}4 = \frac{\left [ 0.4 \right ]^{x}}{\left [ 0.1 \right ]^{x}}\end{array} \)
\(\begin{array}{l}4 = \left (\frac{0.4}{0.1 } \right )^{x}\end{array} \)
\(\begin{array}{l}\left (4 \right )^{1} = \left (4 \right )^{x}\end{array} \)
x = 1
Dividing equation (3) by (2), we get
\(\begin{array}{l}\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{k\left [ 0.3 \right ]^{x} \left [ 0.4 \right ]^{y}}{k\left [ 0.3 \right ]^{x} \left [ 0.2 \right ]^{y}}\end{array} \)
\(\begin{array}{l}4 = \left (\frac{0.4}{0.2} \right )^{y}\end{array} \)
\(\begin{array}{l}4 = 2^{y}\end{array} \)
\(\begin{array}{l}2^{2} = 2^{y}\end{array} \)
y = 2
Hence, the rate law is
Rate =
\(\begin{array}{l}k \left [ A \right ] \left [ B \right ]^{2}\end{array} \)
\(\begin{array}{l}k = \frac{Rate}{\left [ A \right ] \left [ B \right ]^{2}}\end{array} \)
From experiment 1, we get
\(\begin{array}{l}k = \frac{6.0 \times 10^{-3} mol\; L^{-1} \;min^{-1}}{\left (0.1 \;mol \;L^{-1} \right ) \left (0.1 \;mol \;L^{-1} \right )^{2}}\end{array} \)
= 6.0
\(\begin{array}{l}L^{2} \;mol^{-2} \;min^{-1}\end{array} \)
From experiment 2, we get
\(\begin{array}{l}k = \frac{7.2 \times 10^{-2} mol\; L^{-1} \;min^{-1}}{\left (0.3 \;mol \;L^{-1} \right ) \left (0.2 \;mol \;L^{-1} \right )^{2}}\end{array} \)
= 6.0
\(\begin{array}{l}L^{2} \;mol^{-2} \;min^{-1}\end{array} \)
From experiment 1, we get
\(\begin{array}{l}k = \frac{2.88 \times 10^{-1} mol\; L^{-1} \;min^{-1}}{\left (0.3 \;mol \;L^{-1} \right ) \left (0.4 \;mol \;L^{-1} \right )^{2}}\end{array} \)
= 6.0
\(\begin{array}{l}L^{2} \;mol^{-2} \;min^{-1}\end{array} \)
From experiment 1, we get
\(\begin{array}{l}k = \frac{2.4 \times 10^{-2} mol\; L^{-1} \;min^{-1}}{\left (0.4 \;mol \;L^{-1} \right ) \left (0.1 \;mol \;L^{-1} \right )^{2}}\end{array} \)
= 6.0
\(\begin{array}{l}L^{2} \;mol^{-2} \;min^{-1}\end{array} \)
Thus, rate constant, k = 6.0
\(\begin{array}{l}L^{2} \;mol^{-2} \;min^{-1}\end{array} \)
Q 12. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table.
| Exp. | \(\begin{array}{l}\frac{A}{mol L^{-1}}\end{array} \) | \(\begin{array}{l}\frac{B}{mol L^{-1}}\end{array} \) | Initial rate \(\begin{array}{l}mol\; L^{-1} \;min^{-1}\end{array} \) |
| 1 | 0.1 | 0.1 | \(\begin{array}{l}2.0 \times 10^{-2}\end{array} \) |
| 2 | — | 0.2 | \(\begin{array}{l}4.0 \times 10^{-2}\end{array} \) |
| 3 | 0.4 | 0.4 | — |
| 4 | — | 0.2 | \(\begin{array}{l}2.0 \times 10^{-2}\end{array} \) |
Ans:
The given reaction is of the first order with respect to A and of zero order with respect to B.
Thus, the rate of the reaction is given by,
Rate =
\(\begin{array}{l}k \left [ A \right ]^{1} \left [ B \right ]^{0}\end{array} \)
Rate =
\(\begin{array}{l}k \left [ A \right ]\end{array} \)
From experiment 1, we get
\(\begin{array}{l}2.0 \times 10^{-2}\, mol\; L^{-1} min^{-1} = k\left ( 0.1 \; mol \; L^{-1} \right ) \\ \\ \Rightarrow k = 0.2\; min^{-1}\end{array} \)
From experiment 2, we get
\(\begin{array}{l}4.0 \times 10^{-2}\, mol\; L^{-1} min^{-1} = 0.2 min^{-1} \left [ A \right ] \\ \\ \Rightarrow \left [ A \right ] = 0.2\; mol\; L^{-1}\end{array} \)
From experiment 3, we get
Rate =
\(\begin{array}{l}0.2 \; min^{-1} \times 0.4 \; mol \; L^{-1}\\ \\ = 0.08\; mol\; L^{-1}min^{-1}\end{array} \)
From experiment 4, we get
\(\begin{array}{l}2.0 \times 10^{-2}\; mol \; L^{-1} min^{-1} = 0.2\; min^{-1} \left [ A \right ] \\ \\ \Rightarrow \left [ A \right ] = 0.1\; mol \; L^{-1}\end{array} \)
Q 13. Calculate the half-life of a first-order reaction from their rate constants given below.
(a)
\(\begin{array}{l}200 \; s^{-1}\end{array} \)
(b)
\(\begin{array}{l}2 \; min^{-1}\end{array} \)
(c)
\(\begin{array}{l}4 \; years^{-1}\end{array} \)
Ans:
(a) Half life,
\(\begin{array}{l}t_{ \frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac {0.693}{200\; s^{-1}} \\ \\ = 3.47\;\times 10^{-3} s\end{array} \)
(Approximately)
(b)
\(\begin{array}{l}t_{ \frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac {0.693}{2\; min^{-1}} \\ \\ = 0.35\; min\end{array} \)
(Approximately)
(c)
\(\begin{array}{l}t_{ \frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac {0.693}{4\; years^{-1}} \\ \\ = 0.173\; years\end{array} \)
(Approximately)
Q 14. The half-life for the radioactive decay of 14C is 5730 years. An archaeological artefact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
Ans:
Here,
\(\begin{array}{l}k = \frac{0.693}{t_{\frac{1}{2}}} \\ \\ = \frac{0.693}{5730}years^{-1}\end{array} \)
It is known that,
\(\begin{array}{l}t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ = \frac{2.303}{0.693/5730} \log \frac{100}{80} \\ \\ = 1845\; years\end{array} \)
(approximately)
Hence, the age of the sample is 1845 years.
Q 15. The experimental data for decomposition of
\(\begin{array}{l}N_{2} O_{5}\end{array} \)
\(\begin{array}{l}\left [2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2} \right ]\end{array} \)
in the gas phase at 318K are given below:
| T(s) | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
| \(\begin{array}{l}10^{2} \times \left [N_{2} O_{5} \right ] mol \; L^{-1}\end{array} \) | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
(a) Plot [N2O5] against t.
(b) Find the half-life period for the reaction.
(c) Draw a graph between log[N2O5] and t.
(d) What is the rate law?
(e) Calculate the rate constant.
(f) Calculate the half-life period from k and compare it with (b).
Ans:
(a)
(b) Time corresponding to the concentration,
\(\begin{array}{l}\frac{1.630 \times 10^{2}}{2}\; mol\; L^{-1} = 81.5 mol\; L^{-1}\end{array} \)
is the half-life. From the graph, the half-life obtained is 1450 s.
(c)
| t(s) | \(\begin{array}{l}10^{2} \times \left [N_{2} O_{5} \right ] mol \; L^{-1}\end{array} \) | \(\begin{array}{l}\log \left [ N_2 O_{5} \right ]\end{array} \) |
| 0 | 1.63 | -1.79 |
| 400 | 1.36 | -1.87 |
| 800 | 1.14 | -1.94 |
| 1200 | 0.93 | -2.03 |
| 1600 | 0.78 | -2.11 |
| 2000 | 0.64 | -2.19 |
| 2400 | 0.53 | -2.28 |
| 2800 | 0.43 | -2.37 |
| 3200 | 0.35 | -2.46 |
(d) The given reaction is of the first order as the plot,
\(\begin{array}{l}\log \left [ N_2 O_{5} \right ]\end{array} \)
v/s t, is a straight line.
Therefore, the rate law of the reaction is
Rate =
\(\begin{array}{l}k \left [ N_2 O_{5} \right ]\end{array} \)
(e) From the plot,
\(\begin{array}{l}\log \left [ N_2 O_{5} \right ]\end{array} \)
v/s t, we obtain
\(\begin{array}{l}Slope = \frac{-2.46 – \left ( -1.79 \right )}{3200 – 0} \\ \\ = \frac{-0.67}{3200}\end{array} \)
Again, the slope of the line of the plot
\(\begin{array}{l}\log \left [ N_2 O_{5} \right ]\end{array} \)
v/s t is given by
\(\begin{array}{l}- \frac{k}{2.303}\end{array} \)
.
Therefore, we obtain
\(\begin{array}{l}- \frac{k}{2.303} = – \frac{0.67}{3200} \\ \\ \Rightarrow k = 4.82 \times 10^{-4}s^{-1}\end{array} \)
(f) Half-life is given by,
\(\begin{array}{l}t_{\frac{1}{2}} = \frac{0.639}{k} \\ \\ = \frac{0.693}{4.82 \times 10^{-4}}S \\ \\ = \frac{1.483}{10^{3}}s \\ \\ = 1438 s\end{array} \)
The value, 1438 s, is very close to the value that was obtained from the graph.
Q 16. The rate constant for a first-order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
Ans:
It is known that,
\(\begin{array}{l}t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ = \frac{2.303}{60\; s^{-1}} \log \frac{1}{1/16} \\ \\ = \frac{2.303}{60\; s^{-1}} \log 16 \\ \\ = 4.6 \times 10^{-2} \left ( approximately \right )\end{array} \)
Hence, the required time is
\(\begin{array}{l}4.6 \times 10^{-2}\; s\end{array} \)
.
Q 17. During the nuclear explosion, one of the products is 90Sr, with a half-life of 28.1 years. If 1µg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
Ans:
\(\begin{array}{l}k = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{28.1}\; y^{-1}\end{array} \)
Here,
It is known that
\(\begin{array}{l}t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ \Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{\left [ R \right ]} \\ \\ \Rightarrow 10 = \frac{2.303}{\frac{0.693}{28.1}}\left ( – \log \left [ R \right ] \right ) \\ \\ \Rightarrow \log \left [ R \right ] = – \frac{10 \times 0.693}{2.303 \times 28.1}\\ \\ \Rightarrow \left [ R \right ] = antilog \left ( – 0.1071 \right )\\ \\ = antiog \left ( 1.8929 \right ) \\ \\ = 0.7814 \mu g\end{array} \)
Therefore,
\(\begin{array}{l}0.7814 \; \mu g\end{array} \)
of
\(\begin{array}{l}^{90}Sr\end{array} \)
will remain after 10 years.
Again,
\(\begin{array}{l}t = \frac{2.303}{k} \;log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ \Rightarrow 60 = \frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{\left [ R \right ]}\\ \\ \Rightarrow log \left [ R \right ] = – \frac{60 \times 0.693}{2.303 \times 28.1}\\ \\ \left [ R \right ] = antilog \; \left ( – 0.6425 \right ) \\ \\ = antilog \left ( 1.3575 \right )\\ \\ = 0.2278 \mu g\end{array} \)
Therefore,
\(\begin{array}{l}0.2278 \mu g \; o\!f \; ^{90}Sr\end{array} \)
will remain after 60 years.
Q 18. For a first-order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of the reaction.
Ans:
For a first-order reaction, the time required for 99% completion is
\(\begin{array}{l}t_{1} = \frac{2.303}{k} \log \frac{100}{100 – 99} \\ \\ = \frac{2.303}{k} \log 100 \\ \\ = 2 \times \frac{2.303}{k}\end{array} \)
For a first-order reaction, the time required for 90% completion is
\(\begin{array}{l}t_{1} = \frac{2.303}{k} \log \frac{100}{100 – 90} \\ \\ = \frac{2.303}{k} \log 10 \\ \\ = \frac{2.303}{k}\end{array} \)
Therefore,
\(\begin{array}{l}t_{1} = 2 \; t_{2}\end{array} \)
Hence, the time required for 99% completion of a first-order reaction is twice the time required for the completion of 90% of the reaction.
Q 19. A first-order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Ans:
For a first-order reaction,
\(\begin{array}{l}t = \frac{2.303}{k} \log \frac{\left [ R \right ]_{0}}{\left [ R \right ]} \\ \\ k = \frac{2.303}{40\; min} \log \frac{100}{100 – 30} \\ \\ = \frac{2.303}{40 \; min} \log \frac{10}{7} \\ \\ = 8.918 \times 10^{-3} \; min^{-1}\end{array} \)
Therefore,
\(\begin{array}{l}t _{\frac{1}{2}}\end{array} \)
of the decomposition reaction is
\(\begin{array}{l}t _{\frac{1}{2}} = \frac{0.693}{k} \\ \\ = \frac{0.693}{8.918 \times 10^{-3}} min \\ \\ = 77.7 \; min \; \left ( approximately \right )\end{array} \)
Q 20. For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.
| t(sec) | P(mm of Hg) |
| 0 | 35.0 |
| 360 | 54.0 |
| 720 | 63.0 |
Calculate the rate constant.
Ans:
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
\(\begin{array}{l}\left ( CH_{3} \right )_{2} CHN \! = \! NCH \left ( CH_{3} \right )_{2 \left ( g \right )} \rightarrow N_{2\left ( g \right )} + C_{6}H_{14\left ( g \right )}\end{array} \)
At t = 0
\(\begin{array}{l}P_{0}\end{array} \)
0 0
At t = t
\(\begin{array}{l}P_{0} – P\end{array} \)
p p
After time, t, total pressure,
\(\begin{array}{l}P_{t} = \left ( P_{0} – p \right ) + p + p \\ \\ \Rightarrow P_{t} = P_{0} + p \\ \\ \Rightarrow p = P_{t} – P_{0} \\ \\ Therefore, P_{0} – p = P_{0} – \left ( P_{t} – P_{0} \right ) \\ \\ = 2P_{0} – P_{t}\end{array} \)
For the first-order reaction,
\(\begin{array}{l}k = \frac{2.303}{t} \log \frac{P_{0}}{P_{0} – p} \\ \\ = \frac{2.303}{t} \log \frac{P_{0}}{2P_{0} – P_{t}}\end{array} \)
When t = 360 s,
\(\begin{array}{l}k = \frac{2.303}{360 \; s} \log \frac{35.0}{2 \times 35.0 – 54.0} \\ \\ = 2.175 \times 10^{-3} s^{-1}\end{array} \)
When t = 720 s,
\(\begin{array}{l}k = \frac{2.303}{720 \; s} \log \frac{35.0}{2 \times 35.0 – 63.0} \\ \\ = 2.235 \times 10^{-3} s^{-1}\end{array} \)
Hence, the average value of the rate constant is.
\(\begin{array}{l}k = \frac{2.21 \times 10^{-3} + 2.235 \times 10^{-3} }{2} \; s^{-1} \\ \\ = 2.21 \times 10^{-3} s^{-1}\end{array} \)
Q 21. The following data were obtained during the first-order thermal decomposition of SO2Cl2
at a constant volume.
\(\begin{array}{l}SO_{2} CL_{2 \left ( g \right )} \rightarrow SO_{2 \left ( g \right )} CL_{2}\left ( g \right )\end{array} \)
| Experiment | Time/s | Total pressure / atm |
| 1 | 0 | 0.5 |
| 2 | 100 | 0.6 |
Calculate the rate of the reaction when the total pressure is 0.65 atm.
Ans:
The thermal decomposition of
\(\begin{array}{l}SO_{2} CL_{2}\end{array} \)
at a constant volume is represented by the following equation.
\(\begin{array}{l}SO_{2} CL_{2 \left ( g \right )} \rightarrow SO_{2 \left ( g \right )} CL_{2}\left ( g \right )\end{array} \)
At t = 0
\(\begin{array}{l}P_{0}\end{array} \)
0 0
At t = t
\(\begin{array}{l}P_{0} – p\end{array} \)
0 0
After time t, total pressure,
\(\begin{array}{l}P_{t} = \left ( P_0 – p \right ) + p + p \\ \\ \Rightarrow P_t = P_0 + p \\ \\ \Rightarrow p = P_t – P_0 \\ \\∴ P_0 – p = P_o – \left ( P_t – P_0 \right ) \\ \\ = 2 P_0 – P_t\end{array} \)
For a first-order reaction,
\(\begin{array}{l}k = \frac{2.303}{t} \log \frac{P_0}{P_0 – p} \\ \\ = \frac{2.303}{t} \log \frac{P_0}{2P_0 – P_t}\end{array} \)
When t = 100s,
\(\begin{array}{l}k = \frac{2.303}{100 \; s} \log \frac{0.5}{2 \times 0.5 – 0.6} \\ \\ = 2.231 \times 10^{-3} \; s^{-1}\end{array} \)
When
\(\begin{array}{l}P_t = 0.65\end{array} \)
atm,
\(\begin{array}{l}P_0 + p = 0.65 \\ \\ \Rightarrow p = 0.65 – P_0 \\ \\ = 0.65 – 0.5 \\ \\ = 0.15 \; atm\end{array} \)
Therefore, when the total pressure is 0.65 atm, the pressure of
\(\begin{array}{l}SO_{2} CL_{2}\end{array} \)
is
\(\begin{array}{l}P_{SOCL_{2} } = P_0 – p \\ \\ = 0.5 – 0.15 \\ \\ = 0.35 \; atm\end{array} \)
Therefore, the rate of the equation, when the total pressure is 0.65 atm, is given by
\(\begin{array}{l}Rate = k \left ( P_{SOCL_{2}} \right ) \\ \\ = \left ( 2.23 \times 10^{-3} s^{-1} \right )\left ( 0.35 \right ) \; atm \\ \\ = 7.8 \times 10^{-4} \; atm \; s^{-1}\end{array} \)
Q 22. The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of the pre-exponential factor?
Ans:
\(\begin{array}{l} k = 2.418 \times 10^{-5} s^{-1}\end{array} \)
T = 546 K
\(\begin{array}{l}E_a = 179.9 kJ\; mol^{-1} = 179.9 \times 10^3\; J\; mol^{-1}\end{array} \)
According to the Arrhenius equation,
\(\begin{array}{l}k =Ae^{-E_a /RT}\\ \\ \Rightarrow I\!n \;k = I\!n A – \frac{E_a}{RT} \\ \\ \Rightarrow \log k = \log A – \frac{E_a}{2.303 \; RT} \\ \\ \Rightarrow \log A = \log k + \frac{E_a}{2.303 \; RT} \\ \\ = \log \left ( 2.418 \times 10^{-5} \; s^{-1} \right ) + \frac{179.9 \times 10^3 J mol^{-1}}{2.303 \times 8.314 \; Jk^{-1} \; mol^{-1} \times 546\; K} \\ \\ = \left ( 0.3835 – 5 \right ) + 17.2082 \\ \\ = 12.5917\end{array} \)
Therefore, A = antilog (12.5917)
\(\begin{array}{l}= 3.9 \times 10^{12 } \; s^{-1}\end{array} \)
(approximately)
Q 23. Consider a certain reaction A → Products with k = 2.0 × 10–2s–1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1.
Ans:
\(\begin{array}{l}k = 2.0 \times 10^{-2} \; s^{-1} \; T = 100 \; s\end{array} \)
\(\begin{array}{l}\left [ A \right ]_0 = 1.0 \; mol^{-1}\end{array} \)
Since the unit k is
\(\begin{array}{l}s^{-1}\end{array} \)
, the given reaction is a first-order reaction.
Therefore,
\(\begin{array}{l}k = \frac{2.303}{t} \log \frac{\left [ A \right ]_0}{\left [ A \right ]} \\ \\ \Rightarrow 2.0 \times 10^{-2}\; s^{-1} = \frac{2.303}{100 \; s} \log \frac{1.0}{\left [ A \right ]} \\ \\ \Rightarrow 2.0 \times 10^{-2} \; s^{-1} = \frac{2.303}{100 \; s}\left ( – \log \left [ A \right ] \right ) \\ \\ \Rightarrow – \log \left [ A \right ] = \frac{2.0 \times 10 ^{-2} \times 100}{2.303} \\ \\ \Rightarrow \left [ A \right ] = antilog \left ( – \frac{2.0 \times 10 ^{-2} \times 100}{2.303} \right ) \\ \\ = 0.135 \; mol \; L^{-1}\end{array} \)
(approximately)
Hence, the remaining concentration of A is
\(\begin{array}{l}0.135 \; mol \; L^{-1}\end{array} \)
Q 24. Sucrose decomposes in an acid solution into glucose and fructose according to the first-order rate law, with t 1/2 = 3.00 hours. What fraction of a sample of sucrose remains after 8 hours?
For the first-order reaction,
\(\begin{array}{l}k = \frac{2.303}{t} \log \frac{\left [ R \right ]_0}{\left [ R \right ]}\end{array} \)
It is given that
\(\begin{array}{l}t_{\frac{1}{2}} = 3.00 \; hours\end{array} \)
.
Therefore,
\(\begin{array}{l}k = \frac{0.693}{t_{\frac{1}{2}}}\end{array} \)
\(\begin{array}{l}\frac{0.693}{3} h^{-1} \\ \\ = 0.231 \; h^{-1}\end{array} \)
Then,
\(\begin{array}{l}0.231 \; h^{-1} = \frac{2.303}{8h} \log \frac{\left [ R \right ]_0}{\left [ R \right ]} \\ \\ \Rightarrow \log \frac{\left [ R \right ]_0}{\left [ R \right ]} = \frac{0.231\; h^{-1} \times 8\; h}{2.303} \\ \\ \Rightarrow \frac{\left [ R \right ]_0}{\left [ R \right ]} = antilog \left ( 0.8024 \right ) \\ \\ \Rightarrow \frac{\left [ R \right ]_0}{\left [ R \right ]} = 6.3445 \\ \\ \Rightarrow \frac{\left [ R \right ]}{\left [ R \right ]_0} = 0.1576\end{array} \)
(approx)
= 0.158
Hence, the fraction of the sample of sucrose that remains after 8 hours is 0.158.
Q 25. The decomposition of hydrocarbon follows the equation
\(\begin{array}{l}k = \left ( 4.5 \times 10_{11} S – 1 \right ) e_{-28000} K/T\end{array} \)
Calculate
\(\begin{array}{l}E_{a}\end{array} \)
.
Ans:
The given equation is
\(\begin{array}{l}k = \left ( 4.5 \times 10_{11} S – 1 \right ) e_{-28000} K/T\end{array} \)
….(i)
The Arrhenius equation is given by
\(\begin{array}{l}k = Ae^{-E_a /RT}\end{array} \)
….(ii)
From equations (i) and (ii), we obtain
\(\begin{array}{l}\frac{E_a }{RT} = \frac{28000\; K}{T} \\ \\ \Rightarrow E_a = R \times 28000 \; K \\ \\ = 8.314\; J \; K^{-1} mol^{-1} \times 28000 \; K \\ \\ 232791\; J \; mol^{-1}\\ \\ = 232.791\; kJ\; mol^{-1}\end{array} \)
Q 26. The rate constant for the first-order decomposition of H2O2 is given by the following equation:
\(\begin{array}{l}log \; k = 14.34 – 1.25 \times 10^4 \; K/T\end{array} \)
Calculate Ea for this reaction, and at what temperature will its half-period be 256 minutes?
Ans:
Arrhenius equation is given by
\(\begin{array}{l}k = Ae^{-E_a/RT} \\ \\ \Rightarrow I\!n\; k = I\!n\; A – \frac{E_a}{RT} \\ \\ \Rightarrow \log k = \log A – \frac{E_a}{2.303\; RT} \;\;\;\;\;\;\;\; . . . (i)\end{array} \)
The given equation is
\(\begin{array}{l}\log k = 14.34 – 1.25 \times 10^{4}\; K/T \;\;\;\;\;\;\;\; . . . (ii) \end{array} \)
From equations (i) and (ii), we obtain
\(\begin{array}{l}\frac{E_a}{2.303\; RT} = \frac{1.25 \times 10^4\; K}{T}\\ \\ \Rightarrow E_a = 1.25 \times 10^4\; K \times 2.303 \times R \\ \\ = 1.25 \times 10^4\; K \times 2.303 \times 8.314 \; J \; K^{-1} mol^{-1} \\ \\ = 239339.3\; J\; mol^{-1}\;\;\;\; (approximately) \\ \\ = 239.34\; kJ\; mol^{-1}\end{array} \)
Also, when
\(\begin{array}{l}t_{\frac{1}{2}} = 256\end{array} \)
minutes,
\(\begin{array}{l}k = \frac{0.693}{t_{\frac{1}{2}}} \\ \\ = \frac{0.693}{256} \\ \\ = 2.707 \times 10^{-3} \; min^{-1}\\ \\ = 4.51 \times 10^{-5}\; s^{-1}\end{array} \)
It is also given that,
\(\begin{array}{l}log\; k = 14.34 – 1.25 \times 10^{4}\; K/T \\ \\ \Rightarrow log \left ( 4.51 \times 10^{-5} \right ) = 14.34 – \frac{1.25 \times 10^4 \; K}{T}\\ \\ \Rightarrow log \left ( 0.654 – 05 \right ) = 14.34 – \frac{1.25 \times 10^4 \; K}{T}\\ \\ \Rightarrow \frac{1.25 \times 10^4\; K}{T} = 18.686 \\ \\ = 668.95\; K \\ \\ = 669\; K\;\;\;\; (approximately)\end{array} \)
Q 27. The decomposition of A into the product has a value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1?
Ans:
From Arrhenius equation, we obtain
\(\begin{array}{l}\log \frac{k_2}{k_1} = \frac{E_a}{2.303\, R}\left ( \frac{T_2 – T_1}{T_1 T_2} \right )\end{array} \)
Also,
\(\begin{array}{l}k_1 = 4.5 \times 10^3\; s^{-1}\end{array} \)
\(\begin{array}{l}T_1 = 273 + 10 = 283\; K \; k_2\\ \\ = 1.5 \times 10^4\; s^{-1}\\ \\ E_a = 60\; kJ\; mol^{-1} = 6.0 \times 10^{4}\; J \; mol^{-1}\end{array} \)
Then,
\(\begin{array}{l}\log \frac{1.5 \times 10^4}{4.5 \times 10^3} = \frac{6.0 \times 10^4\; J\; mol^{-1}}{2.303 \times 8.314\; J\; K^{-1}\; mol^{-1}}\left ( \frac{T_2 – 283}{283\; T_2} \right )\\ \\ \Rightarrow 0.5229 = 3133.627 \left ( \frac{T_2 – 283}{283\; T_2} \right ) \\ \\ \Rightarrow \frac{0.5229 \times 283\; T_2}{3133.627} = T_2 – 283 \\ \\ \Rightarrow 0.9528 \;T_2 = 283\\ \\ \Rightarrow T_2 = 297.019\; K \;\;\;\;\; (approximately) \\ \\ = 297 \; K \\ \\ = 24^{\circ}\end{array} \)
Q 28. The time required for 10% completion of a first-order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s –1. Calculate k at 318K and Ea.
Ans:
For a first-order reaction,
\(\begin{array}{l}t = \frac{2.303}{k} \log \frac{a}{a – x}\\ \\ at \; 298\; K, \; t = \frac{2.303}{k} \log \frac{100}{90}\\ \\ = \frac{0.1054}{k}\\ \\ at\; 308\; K, \; t’ = \frac{2.303}{k’} \log \frac{100}{75} \\ \\ = \frac{2.2877}{k’}\end{array} \)
According to the question,
\(\begin{array}{l}t = t’ \\ \\ \Rightarrow \frac{0.1054}{k} = \frac{0.2877}{k’} \\ \\ \Rightarrow \frac{k’}{k} = 2.7296\end{array} \)
From the Arrhenius equation, we get
\(\begin{array}{l}\log \frac{k’}{k} = \frac{E_a}{2.303\, R}\left ( \frac{T’ – T}{TT’} \right ) \\ \\ \Rightarrow \log \left ( 2.7296 \right ) = \frac{E_a}{2.303 \times 8.314}\left ( \frac{308 – 298}{298 \times 308} \right ) \\ \\ \Rightarrow E_a = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log \left ( 2.7296 \right )}{308 – 298} \\ \\ = 76640.096\; J \, mol^{-1} \\ \\ = 76.64\; kJ\; mol^{-1}\end{array} \)
To calculate k at 318 K,
It is given that,
\(\begin{array}{l}A = 4 \times 10^{10} s^{-1}\end{array} \)
, T = 318 K
Again, from Arrhenius equation, we get
\(\begin{array}{l}\log k = \log A – \frac{E_a}{2.303\; R\, T} \\ \\ = \log \left ( 4 \times 10^{10} \right ) – \frac{76.64 \times 10^{3}}{2.303 \times 8.314 \times 318} \\ \\ = \left ( 0.6021 + 10 \right ) – 12.5876 \\ \\ = -1.9855\end{array} \)
Therefore, k = Antilog(-1.9855)
\(\begin{array}{l}= 1.034 \times 10^{-2} \; s^{-1}\end{array} \)
Q 29. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature.
Ans:
From Arrhenius equation, we get
\(\begin{array}{l}\log \frac{k_2}{k_1} = \frac{E_a}{2.303\, R}\left ( \frac{T_2 – T_1}{T_1 T_2} \right )\end{array} \)
From the question we have,
\(\begin{array}{l}K_2 = 4 K_1\end{array} \)
\(\begin{array}{l}\\ T_1 = 293\; K \\ \\ T_2 = 313\; K\end{array} \)
Therefore,
\(\begin{array}{l}\log \frac{4 K_1}{K_2} = \frac{E_a}{2.303 \times 8.314}\left ( \frac{313 – 293}{293 \times 313} \right )\\ \\ \Rightarrow 0.6021 = \frac{20 \times E_a}{2.303 \times 8.314 \times 293 \times 313} \\ \\ \Rightarrow E_a = \frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}\\ \\ = 52863.33\; J\; mol^{-1}\\ \\ = 52.86\; kJ\; mol^{-1}\end{array} \)
Hence, the required energy of activation is
\(\begin{array}{l}52.86\; kJ\; mol^{-1}\end{array} \)
.
Students will be able to understand the rate of chemical reaction, temperature dependence on the rate of reaction, the Arrhenius equation and the collision theory of chemical reaction. The chapter given here is to assist the students in understanding the lesson in an easy and interesting way. The Class 12 NCERT Solutions for Chemistry Chapter 4 are created by subject experts as per the latest CBSE Syllabus. Students must practise the solutions regularly to prepare effectively for their board examinations.
Class 12 NCERT Solutions for Chemical Kinetics
Chemical Kinetics is a branch of chemistry. It deals with the rate of chemical reaction, the factors affecting it, and the mechanism of the reaction. Based on the rate of reaction, we have 3 types: Instantaneous reactions, Slow reactions and moderately slow reactions. Any chemical reaction that completes in less than 1ps time is called a fast reaction. Any chemical reaction happening for some minutes to some years is called a slow reaction. Intermediate chemical reactions that occur between fast and slow chemical reactions are called moderately slow reactions. This was brief on Chemical Kinetics. Become well-versed in these concepts by referring to NCERT Solutions.
Chemical Kinetics of Class 12 Chemistry, is designed as per the latest CBSE Syllabus
Subtopics for Class 12 Chemistry Chapter – Chemical Kinetics
- The rate of a Chemical Reaction
- Factors Influencing the Rate of a Reaction
- Dependence of Rate on Concentration
- Rate Expression and Rate Constant
- Order of a Reaction
- Molecularity of a Reaction
- Integrated Rate Equations
- Zero Order Reactions
- First-Order Reactions
- Half-Life of a Reaction
- Pseudo-First-Order Reaction
- Temperature Dependence of the Rate of a Reaction
- Effect of Catalyst Ex 4.6 – Collision Theory of Chemical Reactions
Chemical Kinetics Class 12 NCERT is basic to many of the concepts that students will study in the future. To avoid difficulty in the future, students should get a thorough understanding of this chapter using the NCERT Solutions and solve many chemical kinetics numerical problems to get well-versed in the formulas, standard values and equations.
Frequently Asked Questions on NCERT Solutions for Class 12 Chemistry Chapter 4
Q1
What are the topics covered in Chapter 4 of NCERT Solutions for Class 12 Chemistry?
The topics covered in Chapter 4 of NCERT Solutions for Class 12 Chemistry are as follows:
1. The rate of a Chemical Reaction
2. Factors Influencing the Rate of a Reaction
3. Integrated Rate Equations
4. Pseudo-First-Order Reaction
5. Temperature Dependence of the Rate of a Reaction
Q2
How can I score good marks in Chapter 4 of NCERT Solutions for Class 12 Chemistry?
By maintaining a proper study schedule with the time allotted for each chapter, students can score good marks in the Class 12 board exams. The solutions contain various chemical equations which can be used to sharpen the problem-solving skills of students. The theoretical concepts must be learnt at first, followed by the application of formulas in the numericals. The NCERT Solutions for Class 12 Chemistry available at ANAND CLASSES (A School Of Competitions) are the best study resources among various online study materials available on different sites.
Q3
Explain the concept of Chemical Kinetics covered in Chapter 4 of NCERT Solutions for Class 12 Chemistry.
Chemical Kinetics is a branch of Chemistry that deals with the rate of chemical reaction, the mechanism of the reaction and the factors affecting it. According to the rate of reaction, they are three types: slow reaction, instantaneous reaction and moderately slow reaction. The reaction which completes in less than 1 ps time is called a fast reaction. The reaction which occurs from some time to some years is called a slow reaction. The reactions which occur between fast and slow chemical reactions are called moderately slow reactions.
NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics are the study materials that will help students in getting tuned in with the concepts involved in chemical kinetics. The NCERT Solutions for Class 12 Chemistry PDF for chemical kinetics are helpful for the students of CBSE Class 12. These NCERT Solutions are prepared by subject experts at ANAND CLASSES (A School Of Competitions) according to the latest CBSE Syllabus in simple language for easy understanding.
Students aspiring to make a career in the medical or engineering field must practise the NCERT Solutions for Class 12 Chemistry to score well in the board exams as well as various competitive entrance exams. Further, these solutions can also assist students in preparing notes of important concepts or formulae.