Question

The order of the gaseous reaction $Ag⟶2Bg+Cg$ is found to be one at the initial pressure $P_0=90$ torr. The total pressure after ten minutes is found to be $180$ torr. Find the value of the rate constant?

• A. $1.15\times 10^{-3}{s}^{-1}$
• B. $2.30\times 10^{-3}{s}^{-1}$
• C. $3.45\times 10^{-3}{s}^{-1}$
• D. $4.60\times 10^{-3}{s}^{-1}$

Answer 1

Answer: (A)

Given, Initial pressure of $\text{A,}$ ${\text{P}}_0=90\text{ torr}$
Let's assume there is a decrease in pressure of $\text{A}$ by ${\text{P}}_0\text{torr}$ after $\text{10}$ mins. Therefore, after $\text{10}$ mins the reaction mixture will look like,
$A\left(g\right)\to 2B\left(g\right)+C\left(g\right)$
$\left({\left(\text{P}\right)}_0-\text{ p}\right)2\text{p}\text{p}$
${\text{P}}_{\text{t}}={\text{P}}_0+2\text{p}=180\text{ torr}$
$\text{p}=45\text{ torr}$
Thus, after $10$ mins, the pressure of $\text{A}$ will be, ${\text{P}}_{\text{A}}={\text{P}}_0-\text{p}=90-45=45\text{torr}$
$\text{k}=\frac{1}{\text{t}}ln\frac{{\text{P}}_0}{{\text{P}}_{\text{A}}}$
$\text{k}=\frac{1}{600}ln\frac{90}{45}$
$=1.15\times 10^{-3}{\text{ s}}^{-1}$