A) \[0.6\,-x\]
B) \[0.6\,-2x\]
C) \[0.6\,-\frac{x}{2}\]
D) None is correct
Correct Answer: B
Solution :
Concentration of \[[N{{H}_{3}}]=\frac{6}{10}=0.6\] \[2N{{H}_{3}}\,\,\,\,\,\,\rightleftarrows \,\,\,\,\,\,{{N}_{2}}+3{{H}_{2}}\] \[0.6\] \[0\] \[0\] initial \[0.6-2x\] \[x\] \[3x\] at equilibrium Then, concentration of NH3 at equilibrium will be \[0-2x\].You need to login to perform this action.
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