Question

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

Answer 1

Answer: 11.05

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,
$k=\frac{2.303}{t}{\log }_{10}\left(\frac{p_i}{p_A}\right)$
$p_i=6.0atm;p_A=6.0-4.0=2.0atm$
$\therefore 0.10=\frac{2.303}{t}{\log }_{10}\left(\frac{6.0}{2.0}\right)$
$\therefore t=23.03\log (3.0)=11.05\min$