Question

The reaction $FCl{O}_2\left(g\right)\to FClO\left(g\right)+O\left(g\right)$ is first order with a rate constant of 6.76 × 10^-4s^-1 at 322°C.

(a) Calculate the half-life of the reaction at 322°C.

(b) If the initial partial pressure of FClO₂ in a container at 322°C is 0.040 atm, how long will it take to fall to 0.010 atm?$FCl{O}_2\left(g\right)\to FClO\left(g\right)+O\left(g\right)$

Short answer

The time taken for the reaction is $2.1\times {10}^3\text{s}$

Step-by-step solution

Half life of the reaction:

a) Half life of the reaction is given by the eqution given below

$t_{\frac{1}{2}}=\frac{0.693}{k}...........\left(1\right)$

Where k is the rate constant of the reaction, $t_{\frac{1}{2}}$is the half life of the reaction substituting $k=6.76\times {10}^{-4}$ in eqaution1

Half life of the reaction at$322°\text{C}$is$1.0\times {10}^{-3}$s

Time of the reaction:

Rate law of the reaction when initial and final pressure are given

$2.303\log \left(\frac{P_0}{P_t}\right)=kt........\left(2\right)$

Where P₀ and P_t are initial and final pressure k is the rate constant of the reaction.

Substituting the values of initial, final pressures in the equation 2

$$\begin{gathered} 2.303\log \left(\frac{0.040}{0.010}\right)=kt........\left(2\right) \\ =6.7\times {10}^{-4}{s}^{-1}\times t \\ t=\left(\frac{2.303}{6.7\times {10}^{-4}}\right)\log 4 \\ t=2.1\times {10}^{3}s \end{gathered}$$

Hence the time taken for the reaction is$2.1\times {10}^{3}s$