Question

At $518°\mathrm{C},$ the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of $363$ $\mathrm{torr}$, was $1.00\mathrm{torr}{\mathrm{s}}^{-1}$when $5\%$ had reacted and $0.5\mathrm{torr}{\mathrm{s}}^{-1}$ when $33\%$ had reacted. The order of the reaction is

• A. $3$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is D

$2$

Answer: (D)

Explanation for correct option:

Step 1:

Assume the order of reaction with respect to acetaldehyde is $\mathrm{x}$

Hence, for acetaldehyde, the rate of reaction:

$\frac{\mathrm{d}\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}{\mathrm{dt}}=\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^{\mathrm{x}}------\left(\mathrm{i}\right)$

Step 2:

From the ideal gas equation

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \mathrm{P}=\mathrm{CRT} \\ \mathrm{C}=\frac{\mathrm{P}}{\mathrm{RT}}-------\left(\mathrm{ii}\right) \end{gathered}$$

Step 3: Putting the value of equation($\mathrm{ii}$) in equation($\mathrm{i}$)

$\frac{-1}{\mathrm{RT}}\frac{\mathrm{d}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}_{\mathrm{p}}}{\mathrm{dt}}=\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{CHO}\right]}^{\mathrm{x}}-------\left(\mathrm{iii}\right)$

Step 3:

Rate=$1\mathrm{T}\mathrm{orr}{\mathrm{s}}^{-1}$ when $5\%$ had reacted

Rate=$0.5\mathrm{orr}{\mathrm{s}}^{-1}$when $33\%$ had reacted

Step 4:

Writing the above mentioned values in equation ($\mathrm{iii}$)

$$\begin{gathered} \therefore \frac{-1}{\mathrm{RT}}.1=\mathrm{k}{[0.95]}^{\mathrm{x}}\dots \dots \dots \dots \left(\mathrm{iv}\right) \\ \frac{-1}{\mathrm{RT}}.0.5=\mathrm{k}{[0.66]}^{\mathrm{x}}\dots \dots \dots \dots \left(\mathrm{v}\right) \end{gathered}$$

Step 5:

By dividing Equation $\left(\mathrm{iv}\right)\&\left(\mathrm{v}\right)$, we get

$$\begin{gathered} 2={\left(1.43\right)}^{\mathrm{x}} \\ \mathbf{\therefore }\mathbf{x}\mathbf{=}\mathbf{2} \end{gathered}$$

Hence option (D) is the correct option.