Question
Derive integrated rate equation for first order reaction containing gaseous components.
✓
Answer
→ Let us consider a typical first order gas phase reaction.
→ A(g) → B(g) + C(g)
→ Let $p_i$ be the initial pressure of $A$ and $p_t$ the total pressure at time ' $t$ '. Integrated rate equation for such a reaction can be derived as
→ Total pressure $p _{ t }= p _{ A }+ p _{ B }+ p _{ C }$ (pressure units)
→ $p _{ A }, p _{ B }$ and $p _{ C }$ are partial pressures of $A , B$ and $C$ respectively. If $x$ atm can be the decrease in pressure of $A$ at time $t$ and one mole each of $B$ and C is being formed. The increase in pressure of B and C will also be X atm each
where, $p_i$ is the initial pressure at time $t=0$
$
\begin{array}{l}
p _{ t }=\left( p _{ i }- x \right)+ x + x = p _{ i }+ x \\
x =\left( p _{ t }- p _{ i }\right)
\end{array}
$
where, $p_A=p_i-x=p_i-\left(p_t-p_i\right)=2 p_i-p_t$
$
\begin{aligned}
k & =\left(\frac{2.303}{ t }\right)\left(\log \frac{ p _{ i }}{ p _{ A }}\right) \\
& =\frac{2.303}{ t } \log \frac{ p _{ i }}{2 p _{ i }- p _{ t }}
\end{aligned}
$
→ A(g) → B(g) + C(g)
→ Let $p_i$ be the initial pressure of $A$ and $p_t$ the total pressure at time ' $t$ '. Integrated rate equation for such a reaction can be derived as
→ Total pressure $p _{ t }= p _{ A }+ p _{ B }+ p _{ C }$ (pressure units)
→ $p _{ A }, p _{ B }$ and $p _{ C }$ are partial pressures of $A , B$ and $C$ respectively. If $x$ atm can be the decrease in pressure of $A$ at time $t$ and one mole each of $B$ and C is being formed. The increase in pressure of B and C will also be X atm each
where, $p_i$ is the initial pressure at time $t=0$
$
\begin{array}{l}
p _{ t }=\left( p _{ i }- x \right)+ x + x = p _{ i }+ x \\
x =\left( p _{ t }- p _{ i }\right)
\end{array}
$
where, $p_A=p_i-x=p_i-\left(p_t-p_i\right)=2 p_i-p_t$
$
\begin{aligned}
k & =\left(\frac{2.303}{ t }\right)\left(\log \frac{ p _{ i }}{ p _{ A }}\right) \\
& =\frac{2.303}{ t } \log \frac{ p _{ i }}{2 p _{ i }- p _{ t }}
\end{aligned}
$
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