Question

In the decomposition of oxalic acid following data were obtained.

Time (second)	0	300	600
Volume of $KMn{O}_4$ used in mL	22.0	17.0	13.4

If reaction obeys 1st order kinetics then determine the rate constant K and half-life period:

• A. $8.43\times {10}^{-4}sec,13.7minute$
• B. $86\times {10}^{-4}sec,134.3minute$
• C. $8.6\times {10}^{-5}sec,1.343minute$
• D. None of these

Answer 1

Answer: (A)

$8.43\times {10}^{-4}sec,13.7minute$

Volume of $KMn{O}_4$ used is directly proportional to the oxalic acid concentration.
For the first order reaction, the integrated rate law expression is
$k=\frac{2.303}{t}log\frac{a}{a-x}$......(1)
For 300 seconds, the rate constant is
$k=\frac{2.303}{300}log\frac{22.0}{17.0}=0.0086/s$
For 600 seconds, the rate constant is
$k=\frac{2.303}{600}log\frac{22.0}{13.4}=0.0083/s$
The average value of k is $\frac{0.0083+0.0086}{2}=0.00843/s=8.43\times {10}^{-4}s$
The half life period is $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{0.00843}=822s$
Convert the half life period from seconds to minutes
$t_{1/2}=\frac{822}{60}=13.7min$