Question

For the gaseous elementary reaction, $A +2 B \xrightarrow{ K } C ; K =10^{-4} bar ^{-2} \sec ^{-1}$ If initial partial pressure of $A$ and $B$ are $500$ bar and $1$ bar respectively, then find time in second to get the partial pressure of $B$ to be $0.1$ bar. Assume $\left(500 \frac{0.1}{2} \approx 500\right)$.

Answer 1

rate $=10^{-4} P_{A} P_{B}^{2}$
rate $=10^{-4} \times 500 \times P_{B}^{2}$
rate $=5 \times 10^{-2} P_{B}^{2}$
or $-\frac{1}{2} \frac{d P_{B}}{d t}=5 \times 10^{-2} P_{B}^{2}$
or $-\frac{d P_{B}}{d t}=10^{-1} P_{B}^{2}$
For $2 nd$ order, $\frac{1}{P_{B}}=\frac{1}{P_{B}^{0}}+k t$
or $\frac{1}{0.1}=\frac{1}{1}+0.1 \times t$
or $10=1+0.1 t$
or $t=90\, \sec$.