A) \[0.3\]
B) \[0.4\]
C) \[0.2\]
D) \[1.8\]
Correct Answer: C
Solution :
\[\underset{(2.0-x)}{\mathop{\underset{2.0}{\mathop{{{N}_{2}}}}\,}}\,+\underset{(0.5-3x)}{\mathop{\underset{0.5}{\mathop{3{{H}_{2}}}}\,}}\,\underset{2x\,\,At\,\,equilibrium}{\mathop{\underset{0.0\,\,\,\,\,\,Initrially}{\mathop{2N{{H}_{3}}}}\,}}\,\] Given, \[2x=0.2\] \[x=0.1\] \[\therefore \]Number of moles of \[{{H}_{2}}\] at equilibrium \[=0.5-3\times 0.1\] \[=0.5-0.3=0.2\]You need to login to perform this action.
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