Question

For the reaction, $A ( g ) \rightarrow B ( g )+ C ( g )$
Rate law is $\frac{-d[A]}{d t}=k[A]$
At the state, pressure is $100 \,mm$ and after $10 \,min$, pressure is $120 \,mm$. Thus, rate constant (in $min ^{-1}$ ) is

• A. $\frac{2.303}{10} \log \frac{5}{4}$
• B. $\frac{2.303}{10} \log 5$
• C. $\frac{2.303}{10} \log \frac{6}{5}$
• D. $\frac{2.303}{10} \log \frac{5}{6}$

Answer 1

Answer: (A)

It is a gaseous phase reaction.



Thus, total pressure at time $t=p_{i}-x+x+x=\left(p_{i}+x\right)$

$\therefore x=\left(p_{t}-p_{i}\right)$

$\therefore \left(p_{i}-x\right)=\left(2 p_{i}-p_{t}\right)$

$\therefore =(200-120)=80 \,mm$

$k=\frac{2.303}{t} \log \frac{p_{i}}{\left(p_{i}-x\right)}$

$=\frac{2.303}{10 min } \log \frac{100}{80}=\left(\frac{2.303}{10} \log \frac{5}{4}\right) \,min ^{-1}$