| For the reaction |
| \[2{{H}_{2}}(g)\,+2NO(g)\,\to \,{{N}_{2}}(g)+2{{H}_{2}}O(g)\] |
| the observed rate expression is, |
| rate = \[~{{k}_{f}}{{[NO]}^{2}}\,[{{H}_{2}}]\]. |
| The rate expression for the reverse reaction is: |
A) \[{{k}_{b}}[{{N}_{2}}]\,{{[{{H}_{2}}O]}^{2}}/[NO]\]
B) \[~{{k}_{b}}\,[{{N}_{2}}]\,[{{H}_{2}}O]\]
C) \[~{{k}_{b}}\,[{{N}_{2}}]\,{{[{{H}_{2}}O]}^{2}}\]
D) \[{{k}_{b}}[{{N}_{2}}]{{[{{H}_{2}}O]}_{2}}/[{{H}_{2}}]\]
Correct Answer: D
Solution :
[d] \[{{k}_{eq}}=\frac{{{k}_{f}}}{{{k}_{b}}}=\frac{[{{N}_{2}}]}{[{{H}_{2}}]}\frac{{{[{{H}_{2}}O]}^{2}}}{{{[NO]}^{2}}}\] Rearranging \[{{k}_{f}}\,{{[NO]}^{2}}[{{H}_{2}}]=\frac{{{K}_{b}}[{{N}_{2}}]{{[{{H}_{2}}O]}^{2}}}{[{{H}_{2}}]}\] on comparing \[{{R}_{f}}\] and \[{{R}_{b}}\] at equilibrium, \[{{R}_{b}}={{k}_{b}}\frac{[{{N}_{2}}]\,{{[{{H}_{2}}O]}^{2}}}{[{{H}_{2}}]}\]
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