Dihydrogen gas is obtained from natural gas by partial oxidation with stream as per the following endothermic reaction.
`CH_(4)(g)+H_(2)O(g)hArrCO(g)+3H_(2)(g)`
a. Write an expression for `K_(p)` for the above reaction.
b. How will the values o `K_(p)` and composition of equilibrium mixture be affected by
(i) increasxing the pressure (ii) increasing the temperature (iii) using a catalyst?

Explanation

`K_(p)=(P_(CO)xxP_(H_(2))^(3))/(P_(CH_(4))xxP_(H_(2)O))`
`underset(1-x)(CH_(4)(g))+underset(1-x)(H_(2)O(g))hArrunderset(x)(CO(g))+underset(3x)(3H_(2))`
Total number of moles `=(1-x)+(1-x)+x+3x=2-2x`
`P_(CH_(4))=(1-x)/(2-2x)xxP, P_(CO)=x/(2-2x)P`
`P_(H_(2)O)=(1-x)/(2-2x)xxP, P_(H_(2))=(3x)/(2-2x)xxP`
Substituting these values in `K_(p)` equation
`K_(p)=((x/(2-2x)P)(x/(2-2x)P)^(3))/(((1-x)/(2-2x)P)((1-x)/(2-2x)P))=(x^(4)P)/((2-2x)^(2))`
(i) In the above equation the pressure term P is in the numerator. When pressure is increassed to maintain the value of `K_(p)` constant the value in denominator should increase. Thus backward reaction takes place decreasing the amount of products.
(ii) Since the reaction is endothermic increasing the temperature favours the forward reaction forming more amounts of products.
(iii) Catalyst does not influence the equilibrium mixture.