Question

A gas phase reaction $R\to A+B+C+D$ follows first order kinetics. The initial pressure was $x_o$, the pressure after 10 minutes was $y$. The correct expression for the rate constant $k$ is :

• A. $k=\frac{2.303}{10}\log \frac{4x_0}{3x_0-y}$
• B. $k=\frac{2.303}{10}\log \frac{2x_0}{3x_0-y}$
• C. $k=\frac{2.303}{10}\log \frac{3x_0}{4x_0-y}$
• D. $k=\frac{2.303}{10}\log \frac{4x_0}{4x_0-y}$

Answer 1

The correct option is C $k=\frac{2.303}{10}\log \frac{3x_0}{4x_0-y}$

The gas phase reaction is given as
$R\to A+B+C+Dt=0x_{0}0000t=t{x}_0-PPPPP\Rightarrow y=x_0-P+4P\Rightarrow y=x_0+3P\Rightarrow P=\frac{y-x_0}{3}\mathrm{Applying}{1}^{\mathrm{st}}\mathrm{order}\mathrm{kinetics},k=\frac{1}{10}\ln \frac{x_0}{x_0-P}k=\frac{2.303}{10}\log \frac{x_0}{x_0-\left(\frac{y-x_0}{3}\right)}k=\frac{2.303}{10}\log \frac{3x_0}{4x_0-y}$