JEE CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Analyzing +4 | -1

3 \mathrm{~mol}

ammonia is heated in presence of

1 \mathrm{~mol}

of neon. At equilibrium, the mole fraction of hydrogen is found to be 0.5. The value of

\left(K_C\right) \times(V)

for

\mathrm{NH}_3(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g})+\frac{3}{2} \mathrm{H}_2(\mathrm{g})

View answer and explanation

Correct option: C

3 \sqrt{3}

Option C (

3 \sqrt{3}

) is correct because, after calculating the equilibrium constant

K_C

based on the given mole fraction of hydrogen, the value of

K_C \times V

is found to be

3 \sqrt{3}

More Questions from Equilibrium

Practice All Chapter MCQs →

\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