| (1) \[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g),\]\[{{K}_{1}}\] |
| (2) \[{{N}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)2NO\left( g \right),\]\[{{K}_{2}}\] |
| (3) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g){{H}_{2}}O(g),\]\[{{K}_{3}}\] |
| The equation for the equilibrium constant of the reaction \[2N{{H}_{3}}(g)+\frac{5}{2}{{O}_{2}}(g)2NO(g)+3{{H}_{2}}O(g),\] |
| \[({{K}_{4}})\] 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 \[{{K}_{4}}=\frac{{{K}_{2}}K_{3}^{3}}{{{K}_{1}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
