| (1) \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g),{{K}_{1}}\] |
| (2) \[{{N}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2NO(g),{{K}_{2}}\] |
| (3) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\rightleftharpoons {{H}_{2}}O(g),{{K}_{3}}\] |
| The equation for the equilibrium constant of the reaction |
| \[2N{{H}_{3}}(g)+\frac{5}{2}{{O}_{2}}(g)\rightleftharpoons 2NO(g)+3{{H}_{2}}O(g),\] |
| \[({{K}_{4}})\] in terms of \[{{K}_{1}},{{K}_{2}}\] and\[~{{K}_{3}}\] is: |
A) \[\frac{{{K}_{1}}.{{K}_{2}}}{{{K}_{3}}}\]
B) \[\frac{{{K}_{1}}.K_{3}^{2}}{{{K}_{2}}}\]
C) \[{{K}_{1}}{{K}_{2}}{{K}_{3}}\]
D) \[\frac{{{K}_{2}}.K_{3}^{3}}{{{K}_{1}}}\]
Correct Answer: D
Solution :
[d] To calculate the value of \[{{K}_{4}}\] in the given equation we should apply: \[eqn.\left( 2 \right)+eqn.\left( 3 \right)\times 3-eqn.\left( 1 \right)\] hence \[{{K}_{4}}=\frac{{{K}_{2}}K_{3}^{3}}{{{K}_{1}}}\]
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