Ammonia under a pressure of $15{\rm{ }}atm$ at $27^\circ C$ is heated to $347^\circ C$ in a closed vessel in the presence of catalyst. Under the conditions, ${\rm{N}}{{\rm{H}}_3}$ is partially decomposed according to the equation, $2{\rm{N}}{{\rm{H}}_3}\leftrightharpoons{{\rm{N}}_2} + 3{{\rm{H}}_2}$. The vessel is such that the volume remains effectively constant whereas pressure increases to $50{\rm{ }}atm$. Calculate the percentage of ${\rm{N}}{{\rm{H}}_3}$ actually decomposed

JEE CHEMISTRY MCQ: Equilibrium

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Ammonia under a pressure of

15\mathrm{ }atm

27^\circ C

is heated to

347^\circ C

in a closed vessel in the presence of catalyst. Under the conditions,

\mathrm{N}{\mathrm{H}_3}

is partially decomposed according to the equation,

2\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\mathrm{N}_2} + 3{\mathrm{H}_2}

. The vessel is such that the volume remains effectively constant whereas pressure increases to

50\mathrm{ }atm

. Calculate the percentage of

\mathrm{N}{\mathrm{H}_3}

actually decomposed

View answer and explanation

Correct option: A

61.3\%

Option A (

61.3\%

) is correct because it is the percentage of ammonia decomposed calculated from the change in pressure and stoichiometry of the reaction.

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\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2