Question

In the decomposition of oxalic acid following data were obtained.

f reaction obeys 1st order kinetics then determine the rate constant K and half-life period:
A. $8.43 \times {10^{ - 4}}\sec$,$13.7{\text{minute}}$
B. $86 \times {10^{ - 4}}\sec$,$134.3{\text{minute}}$
C. $8.6 \times {10^{ - 5}}\sec$,$1.343{\text{minute}}$
D. None of these

Answer 1

We can calculate the rate constant of the reaction by taking the average rate constant of rate calculated for 300 seconds and rate calculated for 600 seconds. We can calculate the half-life using rate constant and the value 0.693 (a constant value).

Complete step by step answer:
The volume of potassium permanganate used is proportional to the concentration of oxalic acid.
We can write the integrated rate law expression for the first order reaction as,
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{V_0}}}{{{V_t}}}} \right)$
Where,
t represents the time taken
${V_0}$ represents the initial volume
${V_t}$ represents the final volume
Let us now calculate the rate constant for$300\sec$.
For 300 sec, time taken is $300\sec$.
Initial volume is $22.0mL$.
Final volume is $17.0mL$.
Let us now substitute these values in the integrated rate law expression.
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{V_0}}}{{{V_t}}}} \right)$
$\Rightarrow$ $k = \dfrac{{2.303}}{{300}}\log \left( {\dfrac{{22.0mL}}{{17.0mL}}} \right)$
$\Rightarrow$ $k = 0.000859/s$
The rate constant for 300 seconds is $0.000859{s^{ - 1}}$.
Similarly, we can now calculate rate constant for $600\sec$.
For 600 sec, time taken is $600\sec$.
Initial volume is $22.0mL$.
Final volume is $13.4mL$.
Let us now substitute these values in the integrated rate law expression.
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{V_0}}}{{{V_t}}}} \right)$
$\Rightarrow$ $k = \dfrac{{2.303}}{{600}}\log \left( {\dfrac{{22.0mL}}{{13.4mL}}} \right)$
$\Rightarrow$ $k = 0.000826/s$
The rate constant for 600 seconds is $0.000826{s^{ - 1}}$.
Let us now take the average value of rate constant
Average rate constant = $\dfrac{{{k_{\left( {{\text{at}}300\,\sec } \right)}} + {k_{\left( {{\text{at}}\,600\,\sec } \right)}}}}{2}$
Average rate constant = $\dfrac{{0.000859{s^{ - 1}} + 0.000826{s^{ - 1}}}}{2}$
Average rate constant = $0.0008425{s^{ - 1}}$
Average rate constant = $0.000843{s^{ - 1}}$
Average rate constant = $8.43 \times {10^{ - 4}}{s^{ - 1}}$
The average rate constant is $8.43 \times {10^{ - 4}}{s^{ - 1}}$.
We can calculate the half-life period using the value 0.693 and the rate constant.
The expression to calculate half-life period is,
${t_{1/2}} = \dfrac{{0.693}}{k}$
The average rate constant is $8.43 \times {10^{ - 4}}{s^{ - 1}}$.
We can substitute the value of rate constant in the expression to calculate the half-life period.
${t_{1/2}} = \dfrac{{0.693}}{k}$
$\Rightarrow$ ${t_{1/2}} = \dfrac{{0.693}}{{0.000843{s^{ - 1}}}}$
$\Rightarrow$ ${t_{1/2}} = 822s$
We can convert the half-life in seconds to minutes.
$822\sec \times \dfrac{{1\min }}{{60\sec }} = 13.7\min$
The half-life of the reaction is $13.7\min$.

So, the correct answer is Option A .

Note:
We also remember that the rate constant of a first order reaction contains time units and not concentration units. From this we can understand that for a first order reaction, the numerical value of k is independent of the unit in which the concentration is expressed. Even if we change the concentration unit, the numerical value of k for a first order reaction would not change.