Question

The order of the gaseous reaction $Ag \longrightarrow 2 Bg + Cg$ 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,
$A\left(g\right) \rightarrow \, 2B\left(g\right)+C \, \left(g\right)$
$\left(\left(\text{P}\right)_{0} - \text{ p}\right) \, \, \, 2 \text{p} \, \text{p}$
$\text{P}_{\text{t}} = \text{P}_{0} + 2 \text{p} = 180 \text{ torr}$
$\text{p} = 45 \text{ torr}$
Thus, after $10$ mins, the pressure of $\text{A}$ will be, $\text{P}_{\text{A}} = \text{P}_{0} - \text{p} \, = \, 90 - 45 = 45 \, \text{torr}$
$\text{k}=\frac{1}{\text{t}}ln\frac{\text{P}_{0}}{\text{P}_{\text{A}}}$
$\text{k}=\frac{1}{600}ln\frac{90}{45}$
$=1.15\times 10^{- 3}\text{ s}^{- 1}$