Question

The decomposition of benzene diazonium chloride

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

Time for the reaction:

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

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

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

Conversion of hours to seaconds

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

 Final pressure  after time t:

Substituting all the values in equation 1

$$\begin{gathered} 2.303\log \left(\frac{{\text{P}}_t}{0.0088}\right)=-\left(4.3\times {10}^{-5}{s}^{-1}\right)\left(36000s\right) \\ \log {\text{P}}_t\text{-log}\left(0.0088\right)\text{=-}\frac{4.3\times 10\times 36000s}{2.303} \\ \log {\text{P}}_t\text{+2.06=-0.672} \\ \log P_t=-2.732 \\ {P}_t={10}^{-2.732} \\ {P}_t=0.0019 \end{gathered}$$

Therfore the partial pressure after 10.0 hours is 0.0019 atm.