Question
Derive the expression of equilibrium constant, $K_C $for the reaction:
$A + B ⇌ C + D$
$A + B ⇌ C + D$
✓
Answer
Consider a hypothetical reversible reaction $A+B \rightleftharpoons C+D$.
Two reactions, namely, forward and reverse reactions occur simultaneously in a reversible chemical reaction. The rate equations for the forward and reverse reactions are:
Rate $_{\text {forward }} \propto[A][B]$
$\therefore$ Rate $_{\text {forward }}= k _{ f }[A][B].......(1)$
$\therefore$ Rate $_{\text {reverse }} \propto[C][D]$
$\therefore$ Rate $_{\text {reverse }}= k _{ r }[C][D] \ldots .(2)$
At equilibrium, the rates of forward and reverse reactions are equal.
Thus,
$\text { Rate }_{\text {forward }}=\text { Rate }_{\text {reverse }}$
$\therefore k _{ f }[ A ][ B ]= k _{ r }[ C ][ D ]$
$\therefore \frac{ k _f}{ k _r}= K _{ C }=\frac{[ C ][ D ]}{[ A ][ B ]} ......(3)$
$K _{ C }$ is called the equilibrium constant.
Two reactions, namely, forward and reverse reactions occur simultaneously in a reversible chemical reaction. The rate equations for the forward and reverse reactions are:
Rate $_{\text {forward }} \propto[A][B]$
$\therefore$ Rate $_{\text {forward }}= k _{ f }[A][B].......(1)$
$\therefore$ Rate $_{\text {reverse }} \propto[C][D]$
$\therefore$ Rate $_{\text {reverse }}= k _{ r }[C][D] \ldots .(2)$
At equilibrium, the rates of forward and reverse reactions are equal.
Thus,
$\text { Rate }_{\text {forward }}=\text { Rate }_{\text {reverse }}$
$\therefore k _{ f }[ A ][ B ]= k _{ r }[ C ][ D ]$
$\therefore \frac{ k _f}{ k _r}= K _{ C }=\frac{[ C ][ D ]}{[ A ][ B ]} ......(3)$
$K _{ C }$ is called the equilibrium constant.
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