Question

The thermal decomposition of Acetaldehyde $\left({\mathrm{CH}}_3\mathrm{CHO}\right)$occurs as: ${\mathrm{CH}}_3\mathrm{CHO}\to {\mathrm{CH}}_4+\mathrm{CO}$. The kinetic data of the thermal decomposition is given below. The rate expression thus can be given as:

The kinetic data of the thermal decomposition.

$\left({\mathrm{CH}}_3\mathrm{CHO}\right)$ $\mathbf{mol}\mathbf{.}{\mathbf{lit}}^{\mathbf{-}\mathbf{1}}$	Rate $\mathbf{(}\mathbf{mol}\mathbf{.}{\mathbf{lit}}^{\mathbf{-}\mathbf{1}}\mathbf{.}{\mathbf{sec}}^{\mathbf{-}\mathbf{1}}\mathbf{)}$
$1.75\times {10}^{-3}$	$2.06\times {10}^{-11}$
$3.5\times {10}^{-3}$	$8.24\times {10}^{-11}$
$7.0\times {10}^{-3}$	$3.30\times {10}^{-10}$

• A. $\mathrm{k}\left[{\mathrm{CH}}_3\mathrm{CHO}\right]$
• B. $\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^{1/2}$
• C. $\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^2$
• D. $\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^3$

Answer 1

The correct option is C

$\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^2$

The explanation for the correct answer

(C) $\mathbf{k}{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}\mathbf{CHO}\mathbf{\right]}}^{\mathbf{2}}$

Rate law

• It is an expression for a chemical reaction that gives a relationship between the concentration of the reactants and the rate of the reaction.
• Here the rate is given by $\mathrm{r}=\mathrm{k}\times {\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^{\mathrm{x}}$

Step$\mathbf{1}$: Equating the Rate of the reaction and concentration of the reactant

• $2.06\times {10}^{-11}=\mathrm{k}\times {\left[1.75\times {10}^{-3}\right]}^{\mathrm{x}}\to \overline{)1}$
• $8.24\times {10}^{-11}=\mathrm{k}\times {\left[3.5\times {10}^{-3}\right]}^{\mathrm{x}}\to \overline{)2}$
• $3.30\times {10}^{-10}=\mathrm{k}\times {\left[7.0\times {10}^{-3}\right]}^{\mathrm{x}}\to \overline{)3}$

Step$\mathbf{2}$: Solving $\overline{)1}\&\overline{)2}$

• $\frac{{\mathrm{Rate}}_1}{{\mathrm{Rate}}_2}=$$\frac{\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}_1^{\mathrm{x}}}{\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}_2^{\mathrm{x}}}$
• $\frac{{\mathrm{Rate}}_1}{{\mathrm{Rate}}_2}=$$\frac{{\left({\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}_1\right)}^{\mathrm{x}}}{{\left({\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}_2\right)}^{\mathrm{x}}}$
• $\frac{2.06\times {10}^{-11}}{8.24\times {10}^{-11}}=\frac{{(1.75\times {10}^{-3})}^{\mathrm{x}}}{{(3.5\times {10}^{-3})}^{\mathrm{x}}}$
• $\frac{2.06}{8.24}={\left(\frac{1.75}{3.5}\right)}^{\mathrm{x}}$
• $\frac{1}{4}={\left(\frac{1}{2}\right)}^{\mathrm{x}}$
• $\mathrm{x}=2$

Then rate expression is given by $\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^2$

Therefore, Option (C) $\mathbf{k}{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}\mathbf{CHO}\mathbf{\right]}}^{\mathbf{2}}$ is correct.