Question

The reaction

${\mathrm{FOCl}}_2\left(\mathrm{g}\right)\to \mathrm{FOCl}\left(\mathrm{g}\right)+\mathrm{O}\left(\mathrm{g}\right)$

is first order with a rate constant of 6.76 × 10^-4s^-121 at 322°C.

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

(b) If the initial partial pressure of FClO₂in a container at 322°C is 0.040 atm, how long will it take to fall to 0.010 atm?

Short answer

a)Half-life of the reaction is $1.0\times {10}^3\mathrm{s}$

b) The reaction takes place at the time$2.1\times {10}^3\mathrm{S}$

Step-by-step solution

Half-life of the reaction

The relation between rate constant and the half-life is given by the equation

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

Where K is the rate constant and is the half-life period of the reaction, substitute rate constant value in equation 1

role="math" localid="1660200867535" $$\begin{gathered} t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{6.76\times {10}^{-4}} \\ {t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\hspace{0.17em}1.0\times {10}^3 \end{gathered}$$

Therefore half-life for the reaction is $1.0\times {10}^3$ s

Total Time of the reaction

Time t can be calculated by the following equation

$2.303\log \left[\frac{{\mathrm{P}}_0}{{\mathrm{P}}_t}\right]=\mathrm{kt}........\left(1\right)$

Where p₀ is the initial pressure, p_t is the final pressure and k is the rate constant of the reaction.

Substituting all below the values in the equation 2

$$\begin{gathered} {\mathrm{P}}_0\hspace{0.17em}=0.04\mathrm{atm} \\ \mathrm{P}1\hspace{0.17em}=0.010\mathrm{atm} \\ \mathrm{K}=6.7\times {10}^{-4}{\mathrm{s}}^{-1} \\ 2.303\log \left[\frac{0.040}{0.010}\right]\hspace{0.17em}=6.7\times {10}^{-4}{\mathrm{s}}^{-1}\times \mathrm{t} \\ \mathrm{t}=\frac{2.303}{6.7\times {10}^{-4}}\log 4 \\ \mathrm{t}=2.1\times {10}^3\mathrm{s} \end{gathered}$$

Therefore the reaction takes place at 2.1x10³s