Question

The decomposition of $C{H}_{3}CHO$ occurs as $C{H}_{3}CHO\left(g\right)\to C{H}_4\left(g\right)+CO\left(g\right)$. Given below is the kinetic data of the decomposition.

$\left[C{H}_{3}CHO\right]$ mol.lit${}^{-1}$	rate (mol.lit${}^{-1}$.sec${}^{-1}$)
1.75 x 10${}^{-3}$	2.06 x 10${}^{-11}$
3.5 x 10${}^{-3}$	8.24 x 10${}^{-11}$
7.0 x 10${}^{-3}$	3.30 x 10${}^{-10}$

The rate expression thus can be given as:

• A. $K\left[C{H}_{3}CHO\right]$
• B. $K[C{H}_{3}CHO{]}^2$
• C. $K[C{H}_{3}CHO{]}^3$
• D. $K[C{H}_{3}CHO{]}^{1/2}$

Answer 1

Answer: (B)

$K[C{H}_{3}CHO{]}^2$

Let us assume the rate is given by $r=k\times [C{H}_{3}CHO{]}^x$.

Now,

$2.06\times {10}^{-11}=k\times [1.75\times {10}^3{]}^x.................(1)$

$8.24\times {10}^{-11}=k\times [3.5\times {10}^3{]}^x...................(2)$

$3.30\times {10}^{-10}=k\times [7.0\times {10}^3{]}^x...................(3)$

By solving any two of the above three equations, we get the value of $x=2$.

So, it is a second order reaction and $r=k[C{H}_{3}CHO{]}^2$.