Question

For the gas phase reaction, ${C_2}{H_4} + {\text{ }}H \rightleftharpoons {C_2}{H_6}\left[ {\Delta H^\circ {\text{ }} = {\text{ }} - 136.8J{\text{ }}mo{l^{ - 1}}} \right]$, is carried out in vessel, the equilibrium concentration of ${C_2}{H_4}$ can be increased by:
A) Increasing the temperature
B) Decreasing the pressure
C) Removing some ${H_2}$
D) Adding some ${C_2}{H_6}$

Answer 1

As we know that in every chemical reaction the chemical equilibrium is the state where both reactants and products are present together in concentrations which have no tendencies to change with interval of time. Later make use of the Van’t hoff equation to know which type of reaction is taking place.

Complete step-by-step answer:
The change in enthalpy $\Delta H = - 136.8{\text{ }}J/mol$
We know that the reaction gets equilibrium when the rate of forward reaction equals the rate of the reverse reaction. We must know that all concentrations of reactant and product are constant at equilibrium.
Therefore, the given gas phase reaction is
${C_2}{H_4} + {H_2} \rightleftarrows {C_2}{H_6}$
We can call the equilibrium constant as${K_{eqm}}$.
Hence, we can defined as follow
${K_{eqm}} = \dfrac{{[{C_2}{H_6}]}}{{[{C_2}{H_4}][{H_2}]}}$
According to the equation,
${K_1} = {K_{eqm}} = \dfrac{{[{C_2}{H_6}]}}{{[{C_2}{H_4}][{H_2}]}}$
Where, K1 is initial concentration of ${C_2}{H_4}$
After, increasing the concentration of${C_2}{H_4}$,
${K_2} = \dfrac{{[{C_2}{H_6}]}}{{[{C_2}{H_4}][{H_2}]}}$
Where, K2 = new equilibrium constant,
Then, ${K_2} < {\text{ }}{K_1}$
We must know that the Van't Hoff equation relates change in equilibrium constant with change in temperature.
Then, applying Van't Hoff equation
$\dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{ - \Delta H}}{R}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$
As${K_2} < {\text{ }}{K_1}$, thus in $\dfrac{{{K_2}}}{{{K_1}}}$ will have negative value, $\Delta H$is negative.
Therefore, $\Delta H < 0$. So it is exothermic reaction.
To make it $\dfrac{{{K_2}}}{{{K_1}}}$ negative, T2 must be greater than T1. So, by increasing temperature increases concentration of ${C_2}{H_4}$.
According to Chatelier's principle in an exothermic reaction, the increase in temperature favors backward reaction. Therefore, concentration of ${C_2}{H_4}$ also increases.
Hence, we can conclude that the concentration of ${C_2}{H_4}$ is increased by increasing the temperature.

Note: For the equilibrium reaction, ratio of the concentrations of different substances towards the concentrations of those on the other side equals a constant similar for that specific reaction. The ratio is always written with the final products over the reactants. Each concentration must be raised to that power of its stoichiometric coefficient in a reaction.
We must remember that a reversible reaction can proceed in both the forward as well as backward directions. Most reactions are theoretically reversible in the closed system, though some can be considered to be irreversible if they work for the formation of reactants or products.