Question

For the gas phase reaction, $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{(g) + }{{\text{H}}_{\text{2}}}(g)\rightleftharpoons \text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{6}}\text{(g) ; }\\!\\![\\!\\!\text{ }\Delta \text{H = }-32.7\text{ kcal }\\!\\!]\\!\\!\text{ }$carried out in the vessel, the equilibrium concentration of $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$can is increased by:
A) Increasing the pressure
B) Increasing the temperature
C) Removing some $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$
D) Adding some $\text{ }{{\text{H}}_{\text{2}}}\text{ }$

Answer 1

According to Le Chatelier’s principle, when a system at the equilibrium is subjected to change in the stress that is concentration, pressure, or temperature, the equilibrium of the system shifts in a direction that reduces the effect or undo the effect of the change. Thus, adjust the stress elements to obtain the maximized product.

Complete step by step solution:
Le Chatelier’s principle is stated as follows,
If equilibrium is subjected to stress, the equilibrium shifts in such a way as to reduce the stress.
The Le Chatelier’s principle can be simplified as when the system at the equilibrium changes concentration, pressure, or temperature, the equilibrium of the system shifts in a direction that minimizes the effect of the change. In other words, equilibrium is adjusted in such a way that the change is nullified.
We have given that, the ethane $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ and the ethane $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$ are in equilibrium. The enthalpy of reaction is equal to $\text{ }\Delta \text{H = }-32.7\text{ kcal }$. We are interested to find out the condition at which we can obtain the maximum concentration of $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ or equilibrium shifts to the left-hand side.
$\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{(g) + }{{\text{H}}_{\text{2}}}(g)\rightleftharpoons \text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{6}}\text{(g) }\Delta \text{H = }-32.7\text{ kcal }$
1. Effect of temperature:
The equilibrium involves two reactions: one favouring product and the other favouring reactant. if one of the reactions is exothermic then the other must be endothermic. we have given that,
$\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{(g) + }{{\text{H}}_{\text{2}}}(g)\rightleftharpoons \text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{6}}\text{(g) }\Delta \text{H = }-32.7\text{ kcal }$
In equilibrium, the ration favouring the product $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$ is accompanied by the evolution of heat i.e. exothermic, then back reaction must be endothermic (absorption of heat). By le Chatelier’s principle, the increase in the temperature will favour the reaction to form $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$.
Thus, the equilibrium concentration of the $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$can be increase by increasing the temperature.
2. Effect of change of pressure:
If the reaction consists of gas, then concentrations of all the components can be altered by changing the pressure. here,1 mole of $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ reacts with the 1 mole of hydrogen to give 1 mole of $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$.
$\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{(g) + }{{\text{H}}_{\text{2}}}(g)\rightleftharpoons \text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{6}}\text{(g) }\Delta \text{H = }-32.7\text{ kcal }$
The forward reaction is accompanied by a decrease in the number of moles. If the pressure is an increase, the volume will decrease and the number of moles per unit volume will increase. However, we are interested in the higher concentration of reactants, thus a decrease in pressure will increase the number of moles of reactant $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$and hydrogen.
Thus, the equilibrium concentration $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ can is increased by decreasing pressure.
3. Effect of change of concentration:
The addition of more hydrogen gas $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$will increase the concentration of $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$. The equilibrium can be sifted by limiting the concentration of the reactant. As we want the required to shift in the backward reaction, removal of hydrogen gas during the reaction does will reduce the interaction between the $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ and hydrogen and hence decrease the concentration of $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$ and maximized the concentration of $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$. Also adding the $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$ will sift the reaction towards the reactant.
Thus, equilibrium concentration $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ can be increase by removing some $\text{ }{{\text{H}}_{\text{2}}}\text{ }$ or add some $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$. The addition of $\text{ }{{\text{H}}_{\text{2}}}\text{ }$and removal of $\text{ }{{\text{C}}_{2}}{{\text{H}}_{6}}\text{ }$ will favour the forward reaction. Thus, option (C) and (D) are not valid.

Hence, (A) and (B) are correct options.

Note: note that, the above reaction of conversion of ethane to ethane is a hydrogenation reaction. Normally hydrogenation reactions are slow, we add metal catalysts to increase the rate of reaction. The metal catalyst does not affect the equilibrium of the reaction as here both forward and backward reactions are sped up by the same factor. Remember that, the negative sign of enthalpy corresponds to the exothermic reaction and positive to the endothermic reaction.