A) \[1.49\times {{10}^{-5}}L\,mo{{l}^{-1}}\]
B) \[2.22\times {{10}^{-10}}L\,mo{{l}^{-1}}\]
C) \[3.86\times {{10}^{-3}}L\,mo{{l}^{-1}}\]
D) Question is incomplete
Correct Answer: C
Solution :
| [c] \[{{N}_{2}}\left( g \right)+3{{H}_{2}}\left( g \right)\rightleftharpoons 2N{{H}_{3}}\left( g \right)\] |
| At equilibrium \[\left( 1-x \right)\] \[\left( 3-3x \right)\] \[2x\] |
| (x=0.0025) |
| Active masses \[\frac{\left( 1-0.0025 \right)}{4}\frac{\left( 3-0.0075 \right)}{4}\frac{0.0050}{4}\] |
| Applying law of mass action, |
| \[{{\text{K}}_{\text{C}}}\text{=}\frac{{{\left[ \text{N}{{\text{H}}_{\text{3}}} \right]}^{\text{2}}}}{\left[ {{\text{N}}_{\text{2}}} \right]{{\left[ {{\text{H}}_{\text{2}}} \right]}^{\text{3}}}}\text{=}\frac{{{\left( \frac{\text{0}\text{.0050}}{\text{4}} \right)}^{\text{2}}}}{\left( \frac{\text{0}\text{.9975}}{\text{4}} \right){{\left( \frac{\text{2}\text{.9925}}{\text{4}} \right)}^{\text{3}}}}\] |
| \[\text{=1}\text{.49 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{ }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{\text{-1}}}\] \[{{K}_{C}}\] for the reaction, |
| \[\frac{1}{2}{{N}_{2}}\left( g \right)+\frac{3}{2}{{H}_{2}}\left( g \right)\rightleftharpoons N{{H}_{3}}\left( g \right).\] |
| Is equal to \[\sqrt{{{K}_{C}}},\] |
| i.e. \[\text{K=}\sqrt{{{\text{K}}_{\text{C}}}}\text{=}\sqrt{\text{1}\text{.49 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}}\text{=3}\text{.86 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{L mo}{{\text{l}}^{\text{-1}}}\] |
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