Question

Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per the following endothermic reaction:
$C{{H}_{4}}(g)+{{H}_{2}}O\left( g \right)\rightleftharpoons CO\left( g \right)+3{{H}_{2}}\left( g \right)$
(a) Write an expression ${{K}_{p}}$for the above reaction.
(b) How will the values of ${{K}_{p}}$ and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature and
(iii) Using a catalyst?

Answer 1

Hint: (a)${{K}_{p}}$ is the equilibrium constant calculated from the partial pressure equation of a reaction. It states the relationship between reactant pressures and product pressures.

(b)The change in the composition of equilibrium and the value of ${{K}_{p}}$can be determined by considering the Le-Chatelier principle.
Considering various conditions of Le-Chatlier's principle we could comment on how the equilibrium will be established and the change in the composition concerning the change in different parameters.

Complete step-by-step answer:(a)In the question, a reaction is given in which by partial oxidation of natural gas hydrogen gas is produced. And the given reaction is a balanced equation, hence we can write the equilibrium constant of partial pressure (${{K}_{p}}$) directly.
To write the equation for${{K}_{p}}$, the partial pressures of the products should be divided by the partial pressure of the reactants.
${{K}_{p}}=\dfrac{{{P}_{CO}}P_{{{H}_{2}}}^{3}}{{{P}_{C{{H}_{4}}}}{{P}_{{{H}_{2}}O}}}$

(b)(i) So the first case is about when the pressure increases what will be the change and in which direction the reaction will proceed to re-establish equilibrium. When the pressure is increased according to Le-Chatelier's principle, the reaction will take place in that direction that has the least or which will produce the least number of moles of a substance. In that case, the reaction will proceed towards the left i.e a backward reaction will be carried out.
(ii) In the second case the temperature increases ie the reaction is endothermic. In an endothermic reaction, the reaction takes place in the forward reaction so that the reactants will absorb the heat and re-ensure the equilibrium of the system.
(iii) There is no effect of a catalyst on equilibrium, catalyst is only used to alter the rate of the reaction. The equilibrium state of the system has no significance for the catalyst.

Note:(a)Always balance the chemical equation before writing the equation for equilibrium constant. If the stoichiometric constant is associated with the atom it should be written as the power of partial pressure of that atom.
The term ${{K}_{C}}$ and ${{K}_{p}}$ are different, both are proportionality constant, or say the equilibrium constant but both represent different parameters, one represents the concentration and the other partial pressure.

(b)The catalyst only alters the rate of the reaction, but by using the catalyst in a reaction we could easily attain the equilibrium state.
If in a system the pressure is decreased then the reaction will happen in that direction in which more moles of substances are produced.
In an exothermic reaction, the direction of reaction happening will be from right to left i.e backward reaction will take place.