question_answer 56)

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

Answer:

(a) \[{{K}_{p}}=\frac{[{{p}_{CO}}]{{[{{p}_{{{H}_{2}}}}]}^{3}}}{[{{p}_{C{{H}_{4}}}}][{{p}_{{{H}_{2}}O}}]}\] (b) (i) \[\Delta {{n}_{g}}=4-2=2\] \[\Delta {{n}_{g}}>0\] The reaction will shift in backward direction (lower volume direction) when pressure is increased. However, \[{{K}_{p}}\]is unaffected by \[{{K}_{p}}\]pressure change. (ii) The reaction is endothermic, hence, the equilibrium will shift in forward direction by rise in temperature. The equilibrium constant \[{{K}_{p}}\] will also increase with rise in temperature. \[{{\log }_{10}}\left( \frac{{{K}_{2}}}{{{K}_{1}}} \right)=\frac{\Delta H}{2.303}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] (iii) Composition of equilibrium mixture and the value of \[{{K}_{p}}\], both are unaffected by the catalyst.