Question

Consider a first order gas phase decomposition reaction given below:
$A _{( g )} \longrightarrow B _{( g )}+ C _{( g )}$
The initial pressure is $6.0 \,atm$. The pressure drops to $3.0 \,atm$ after $6.93 \min$. How much time (in minutes) would it take to lower the partial pressure of $A_{(g)}$ by $4.0\, atm$ ?
[Consider: $\log _{10}(3)=0.48$ ]

Answer 1

Since the partial pressure reduces to half after $6.93$ minutes,
the half-life, $t _{1 / 2}=6.93 \min$
$k =\frac{0.693}{ t _{1 / 2}}=\frac{0.693}{6.93}=0.10 \min ^{-1}$
For a gas phase first order reaction,
$k =\frac{2.303}{ t } \log _{10}\left(\frac{ p _{ i }}{ p _{ A }}\right)$
$p _{ i }=6.0 \,atm ; p _{ A }=6.0-4.0=2.0 \,atm$
$\therefore 0.10=\frac{2.303}{ t } \log _{10}\left(\frac{6.0}{2.0}\right)$
$\therefore t =23.03 \log (3.0)=11.05 \min$