JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

For the given reaction,

2 \mathrm{~A}(\mathrm{~s})+\mathrm{B}(\mathrm{g})=\mathrm{C}(\mathrm{g})+2 \mathrm{D}(\mathrm{s})+\mathrm{E}(\mathrm{s})

the degree of dissociation of

\mathrm{B}

was found to be

20 \%

300 \mathrm{~K}

and

24 \%

500 \mathrm{~K}

. The rate of forward reaction

View answer and explanation

Correct option: C

depends on temperature only and increases with increase in temperature.

Option C is correct because, with

\Delta n_g = 0

, pressure has no effect, and the forward reaction rate depends only on temperature, increasing with temperature due to the endothermic nature.

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