Question

The decomposition of benzene diazonium chloride

${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}{\mathbf{\text{N}}}_{\mathbf{2}}\mathbf{\text{Cl}}\mathbf{\to }{\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}\mathbf{\text{Cl}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{\text{N}}}_{\mathbf{2}}$

follows first-order kinetics with a rate constant of 4.3 × 10 ^-5 s^-1 at 20⁰ C . If the initial partial pressure of C₆H₅N₂Cl is 0.0088 atm, calculate its partial pressure after 10.0 hour.

Short answer

Partial pressure after 10.0 hours is found to be 0.0019 atm.

Step-by-step solution

Step-1: Time for the reaction.

Rate law or first order reaction when initial and final pressures are given

$2.303\log \left(\frac{P_0}{P_t}\right)=kt.......\left(1\right)$

Where $P_0\text{and}{P}_t$are initial and final pressure k is the rate constant of the reaction.

Conversion of hours to seconds

$\begin{array}{c}t=10.0hour\\ t=10.0hour\times \frac{60\min }{1hour}\times \frac{60s}{1\min }\\ t=36000s\end{array}$

Step-2: Final pressure after time t.

Substituting all the values in equation 1

$\begin{array}{c}2.303\log \left(\frac{P_t}{0.0088}\right)=-\left(4.3\times {10}^{-5}{s}^{-1}\right)\left(36000s\right)\\ \log pt-\log \left(0.0088\right)=-\frac{4.3\times 10\times 36000s}{2.303}\\ \log p_t+2.06=-0.672\\ \log p_t=-2.732\\ P_t={10}^{-2.732}\\ P_t=0.0019\end{array}$

Therefore the partial pressure after 10.0 hours is 0.0019 atm.