Question

The below given first-order reaction has initial and final moles of ${\mathrm{PCl}}_5$ as $50$ and $10$ respectively in $120$minutes. Find the rate constant. Given ${\log }_{10}5=0.693$

Answer 1

Step 1: Given data

Initial moles of ${\mathrm{PCl}}_5=50$

Final moles of ${\mathrm{PCl}}_5=10$

Time$=120\min$

Step 2: Applying the formula

$\underset{\mathrm{Phosphorous}\mathrm{pentachloride}}{{\mathrm{PCl}}_5\left(\mathrm{g}\right)}\to \underset{\mathrm{Phosphorous}\mathrm{trichloride}}{{\mathrm{PCl}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Chlorine}\mathrm{gas}}{{\mathrm{Cl}}_2\left(\mathrm{g}\right)}$

The given reaction is first order.

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\log \left[\frac{{\mathrm{A}}_0}{\mathrm{At}}\right]$

where ${\mathrm{A}}_{°}$ is the initial concentration and ${\mathrm{A}}_{\mathrm{t}}$ is concentration after some time, Putting the values we get,

$\mathrm{K}=\frac{2.303}{120}\log \frac{50}{10}$

$\mathrm{K}=\frac{2.303}{120}\log 5$

$\mathrm{K}=\frac{2.303}{120}\times 0.693$

$\mathrm{K}=0.01329{\min }^{-1}$ or $1.32\times {10}^{-2}{\min }^{-1}$

Therefore, the rate constant of the given equation is $1.32\times {10}^{-2}{\min }^{-1}$