JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

A vessel contains

\mathrm{A}(\mathrm{g})

and

\mathrm{B}(\mathrm{g})

2 \mathrm{~atm}

and

4 \mathrm{~atm}

respectively at

\mathrm{TK}

, the mixture is allowed to attain equilibrium at

\mathrm{TK}

, according to the reaction

8 \mathrm{B}(\mathrm{g}) \rightleftharpoons 8 \mathrm{A}(\mathrm{g})+\mathrm{C}(\mathrm{s})

At equilibrium,

\left(\frac{\mathrm{n_A}}{\mathrm{n_B}}\right)_{\text {eq }}=\left(\frac{\mathrm{n_B}}{\mathrm{n_A}}\right)_{\text {initial }}

Find the value of

\mathrm{y}

\mathrm{K_c}=2^\mathrm{y}

View answer and explanation

Correct option: A

Option A (8) is correct because the equilibrium constant Kc is calculated to be 2⁸, thus y=8.

More Questions from Equilibrium

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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