Question
Derive integrated rate equation for first order reaction containing gaseous components.
✓
Answer
$\rightarrow$ Let us consider a typical first order gas phase reaction.
$\rightarrow \ce{A(g) \rightarrow B(g) + C(g)}$
$\rightarrow$ 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
$\rightarrow$ Total pressure $p _{ t }= p _{ A }+ p _{ B }+ p _{ C } ($pressure units$)$
$\rightarrow p _{ A }, p _{ B }$ 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 $t=0$
$p _{ t }=\left( p _{ i }- x \right)+ x + x = p _{ i }+ x$
$x =\left( p _{ t }- p _{ i }\right)$
where, $p_A=p_i-x=p_i-\left(p_t-p_i\right)=2 p_i-p_t$
$k =\left(\frac{2.303}{ t }\right)\left(\log \frac{ p _{ i }}{ p _{ A }}\right)$
$ =\frac{2.303}{ t } \log \frac{ p _{ i }}{2 p _{ i }- p _{ t }}$
$\rightarrow \ce{A(g) \rightarrow B(g) + C(g)}$
$\rightarrow$ 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
$\rightarrow$ Total pressure $p _{ t }= p _{ A }+ p _{ B }+ p _{ C } ($pressure units$)$
$\rightarrow p _{ A }, p _{ B }$ 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 $t=0$
$p _{ t }=\left( p _{ i }- x \right)+ x + x = p _{ i }+ x$
$x =\left( p _{ t }- p _{ i }\right)$
where, $p_A=p_i-x=p_i-\left(p_t-p_i\right)=2 p_i-p_t$
$k =\left(\frac{2.303}{ t }\right)\left(\log \frac{ p _{ i }}{ p _{ A }}\right)$
$ =\frac{2.303}{ t } \log \frac{ p _{ i }}{2 p _{ i }- p _{ t }}$
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