Question

In a certain gaseous reaction $A \longrightarrow B$, the initial pressure is $214 \,atm$ and the rate constant is $2.303 \times 10^{-4} s ^{-1}$. What would be pressure (in $atm )$ of $A$ after $5$ mins?
[Given: $10^{0.03}=1.07$ ]

Answer 1

Initial pressure $P_{i}=214$ atm
Rate constant $k =2.303 \times 10^{-4} s ^{-1}$
Unit of rate constant indicates first order reaction.
Time, $t =5$ minutes $=5 \times 60\, s$
The integrated rate law is,
$k =\frac{2.303}{ t } \log _{10}\left(\frac{ P _{ i }}{ P _{ f }}\right) $
$2.303 \times 10^{-4}=\frac{2.303}{5 \times 60} \log _{10}\left(\frac{214}{ P _{ f }}\right) $
$\therefore 0.03=\log _{10}\left(\frac{214}{ P _{ f }}\right) $
$\therefore 1.07=\frac{214}{ P _{ f }} $
$\therefore P _{ f }=\frac{214}{1.07}=200 \,atm$