Question

Two moles of $PC{l_5}$ were heated to $327^\circ C$ in a closed two litre vessel and when equilibrium was achieved, $PC{l_5}$ was found to be $40\%$ dissociated into $PC{l_3}$ and $C{l_2}$. Calculate the equilibrium constant ($K_c$) for the reaction.

Answer 1

We know that the equilibrium constant of a chemical reaction gives understanding into the connection between the products and reactants when a substance response arrives at balance. For instance, the harmony steady of fixation (indicated by $K_c$) of a compound response at balance can be characterized as the proportion of the convergence of items to the centralization of the reactants, each raised to their separate stoichiometric coefficients.

Complete answer:
Measure of the responding species and the product can be determined and afterward apply the $40\%$ separation. Appropriately address $K_p$ and $K_c$ by utilizing $K_p{\text{ }} = {\text{ }}K_c{\text{ }}{\left( {RT} \right)^{\Delta n}}$ Reaction for separation of$PC{l_5}$.
The dissociation equation is,
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
Given:
Measure of $PC{l_5}$ (at first) = $2$ moles
Percentage separation at equilibrium = $40\%$
The number of moles of $PC{l_5}$ separated at equilibrium is calculated as,
The number of moles of $PC{l_5} = \dfrac{{40}}{{100}} \times 2 = 0.8mole$
The quantity of $PC{l_5}$, $PC{l_3}$ and $C{l_2}$ at equilibrium is $2 - 0.8 = 1.2$
The concentrations of the species in the reactant and product side is calculated as,
$\left[ {PC{l_5}} \right] = \dfrac{{1.2}}{2} = 0.6mol{L^{ - 1}}$
$\left[ {PC{l_3}} \right] = \dfrac{{0.8}}{2} = 0.4mol{L^{ - 1}}$
$\left[ {C{l_2}} \right] = \dfrac{{0.8}}{2} = 0.4mol{L^{ - 1}}$
The equilibrium constant ($K_c$) is calculated as follows,
${K_c} = \dfrac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} = \dfrac{{0.4mol{L^{ - 1}}\left( {0.4mol{L^{ - 1}}} \right)}}{{\left( {0.6mol{L^{ - 1}}} \right)}}$
On simplification we get,
${K_c} = 0.267mol{L^{ - 1}}$

Note:
We have to remember that the equilibrium constant of a chemical reaction can be characterized as the proportion between the measure of reactant and the measure of product which is utilized to decide compound conduct. At equilibrium, Rate of the forward response is equal to the rate of the regressive reaction. At a specific temperature, the rate constants are consistent. The proportion of the rate steady of forward response to the rate constant of in reverse response ought to be a steady and is called a balance consistent $\left( {{K_{equ}}} \right).$K is the proportion of the general measure of items to reactants at balance while Q is the proportion anytime of season of the response. The Q worth can measure up to K to decide the bearing of the response to occur.