# Solve this

Question:

$\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$

In the above first order reaction the concentration of $\mathrm{PCl}_{5}$ reduces from initial concentration $50 \mathrm{~mol} \mathrm{~L}^{-1}$ to $10 \mathrm{~mol} \mathrm{~L}^{-1}$ in 120 minutes at $300 \mathrm{~K}$. The rate constant for the reaction at $300 \mathrm{~K}$ is $\mathrm{x} \times 10^{-2} \mathrm{~min}^{-1}$. The value of $\mathrm{x}$ is $[$ Given $\log 5=0.6989]$

Solution:

$\mathrm{PCl}_{5(\mathrm{~g})} \frac{\mathrm{I} \text { order }}{300 \mathrm{~K}} \longrightarrow \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}$

$\mathrm{t}=0 \quad 50 \mathrm{M}$

$\mathrm{t}=120 \mathrm{~min} 10 \mathrm{M}$

$\Rightarrow \mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{0}\right]}{\left[\mathrm{A}_{\mathrm{t}}\right]}$

$\Rightarrow \mathrm{K}=\frac{2.303}{120} \log \frac{50}{10}$

$\Rightarrow \mathrm{K}=\frac{2.303}{120} \times 0.6989=0.013413 \mathrm{~min}^{-1}$

$=1.3413 \times 10^{-2} \mathrm{~min}^{-1}$

$1.34 \Rightarrow$ Nearest integer $=1$