Chemical Equilibrium
4.0 Equilibrium Constant
4.0 Equilibrium Constant
It is proposed by Goldberg and Wage. According to this law, rate of reaction at a given temperature is directly proportional to the active masses of reactants raised to the power equal to stoichiometric coefficient of the balanced chemical equation at particular instant of reaction
$$aA + bB \rightleftharpoons cC + dD$$
$$rate\ of\ forward\ reaction \propto {\left[ A \right]^a}{\left[ B \right]^b}$$
$$rate\ of\ forward\ reaction = {K_f}{\left[ A \right]^a}.{\left[ B \right]^b}$$
Similarly, $$rate\ of\ backward\ reaction \propto {\left[ C \right]^c}{\left[ D \right]^d}$$
$$rate\ of\ backward\ reaction = {K_b}{\left[ C \right]^c}.{\left[ D \right]^d}$$
where $K_f$ and $K_r$ are the forward and backward rate constant or velocity constant respectively of the reaction at that temperature.
At equilibrium, the rate of both forward and backward reactions become equal and after achieving equilibrium, the concentration of reactants and product remain constant i.e.,
$$\begin{equation} \begin{aligned} {K_f}\left[ A \right]\left[ B \right] = {K_b}\left[ C \right]\left[ D \right] \\ \Rightarrow {K_c} = \frac{{{K_f}}}{{{K_b}}} = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}} \\\end{aligned} \end{equation} $$
where ${K_c}$ is the equilibrium constant and $\left[ {\ } \right]$ denotes active masses.
Unit of rate constant is given as
$$\left( K \right) = {\left[ {moles/litre} \right]^{1 - n}}tim{e^{ - 1}}$$
where $n$ is order of reaction.
Example 1. For an equilibrium reaction, the rate constants for the forward and the backward reaction are $2.40 \times {10^{ - 4}}$ and $8.00 \times {10^{ - 5}}$ respectively. Calculate the equilibrium constant for the reaction.
Solution: Equilibrium constant is calculated as $$K = \frac{{{k_f}}}{{{k_b}}} = \frac{{2.40 \times {{10}^{ - 4}}}}{{8.00 \times {{10}^{ - 5}}}} = 3$$
Example 2. Calculate the value of equilibrium constant ${K_c}$ for the system below, if ${10}$ moles of ${N_2}$, ${4}$ moles of ${H_2}$ and ${8}$ moles of $N{H_3}$ were present in $2.0L$ reaction vessel at equilibrium. $${N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right)$$
Solution: Concentration of different species is calculated as
$$\left[ {{N_2}} \right] = \frac{{10mol}}{{2L}} = 5mol{L^{ - 1}}$$
$$\left[ {{H_2}} \right] = \frac{{4mol}}{{2L}} = 2mol{L^{ - 1}}$$
$$\left[ {N{H_3}} \right] = \frac{{8mol}}{{2L}} = 4mol{L^{ - 1}}$$
$${K_C} = \frac{{{{\left[ {N{H_3}} \right]}^2}}}{{{{\left[ {{H_2}} \right]}^3}\left[ {{N_2}} \right]}} = \frac{{{{\left( 4 \right)}^2}}}{{{{\left( 2 \right)}^3}\left( 5 \right)}} = 0.4$$
Example 3. At ${500^ \circ }C$, the equilibrium constant for the reaction $${N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right)$$ is $6.02 \times {10^{ - 2}}mo{l^{ - 2}}{L^2}$. What is the value of ${K_P}$ at the same temperature?
Solution: $${K_P} = {K_c}{\left( {RT} \right)^{\Delta n}}$$
$${K_P} = \left( {6.02 \times {{10}^{ - 2}}mo{l^{ - 2}}{L^2}} \right) \times {\left( {0.082\ Latm{K^{ - 1}}mo{l^{ - 1}} \times 773K} \right)^{ - 2}}$$
$${K_P} = 1.5 \times {10^{^{ - 5}}}at{m^{ - 2}}$$
