In the reversible reaction shown below, $k_f$ is the forward reaction rate constant and $k_r$ is the reverse reaction rate constant.
The CORRECT expression representing the reaction equilibrium constant is ______.
\[A + B \underset{k_r}{\overset{k_f}{\rightleftharpoons}} C + D\]

Question

Consider the equilibrium reactions:

$H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-$ $(K_1)$
$H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$ $(K_2)$
$HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}$ $(K_3)$

The equilibrium constant, $K_c$, for the following dissociation:

$H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-}$

is:

Answer 1

This also follows directly from the standard thermodynamic definition of the equilibrium constant, without writing out rate laws first. For a general reversible reaction, the equilibrium constant is defined as the ratio of the rate constant of the forward process to the rate constant of the reverse process, precisely because it is these two constants that set how far the reaction shifts before settling down.

Physically, a large $k_f$ relative to $k_r$ means the forward reaction proceeds much faster than the reverse one, so the system settles with more product ($C$, $D$) than reactant ($A$, $B$) present, corresponding to a large value of $K$. Conversely, if $k_r$ dominates over $k_f$, the system favors the reactants and $K$ should come out small.

Only the ratio $\dfrac{k_f}{k_r}$ behaves this way. The reciprocal $\dfrac{k_r}{k_f}$ would behave backwards, and the sum $k_f+k_r$ or product $k_f \times k_r$ do not correspond to any meaningful equilibrium quantity at all.

\[\boxed{K = \dfrac{k_f}{k_r}}\]

Answer 2

This also follows directly from the standard thermodynamic definition of the equilibrium constant, without writing out rate laws first. For a general reversible reaction, the equilibrium constant is defined as the ratio of the rate constant of the forward process to the rate constant of the reverse process, precisely because it is these two constants that set how far the reaction shifts before settling down.

Physically, a large $k_f$ relative to $k_r$ means the forward reaction proceeds much faster than the reverse one, so the system settles with more product ($C$, $D$) than reactant ($A$, $B$) present, corresponding to a large value of $K$. Conversely, if $k_r$ dominates over $k_f$, the system favors the reactants and $K$ should come out small.

Only the ratio $\dfrac{k_f}{k_r}$ behaves this way. The reciprocal $\dfrac{k_r}{k_f}$ would behave backwards, and the sum $k_f+k_r$ or product $k_f \times k_r$ do not correspond to any meaningful equilibrium quantity at all.

\[\boxed{K = \dfrac{k_f}{k_r}}\]

In the reversible reaction shown below, \(k_f\) is the forward reaction rate constant and \(k_r\) is the reverse reaction rate constant.
The CORRECT expression representing the reaction equilibrium constant is ______.
\[A + B \underset{k_r}{\overset{k_f}{\rightleftharpoons}} C + D\]

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Chemical equilibria

Questions Asked in GATE ES exam

In the reversible reaction shown below, \(k_f\) is the forward reaction rate constant and \(k_r\) is the reverse reaction rate constant.The CORRECT expression representing the reaction equilibrium constant is ______.\[A + B \underset{k_r}{\overset{k_f}{\rightleftharpoons}} C + D\]