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JEE CHEMISTRY MCQ: Equilibrium

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STATEMENT-1 : The equilibrium constant of the exothermic reaction at high temperature decreases.

STATEMENT-2 : Since

\ln \frac{K_2}{K_1}=\frac{\Delta H^{\circ}}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]

and for exothermic reaction,

\Delta H^{\circ}=-

ve and thereby;

\frac{K_2}{K_1} < 1

View answer and explanation

Correct option: A

If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1

Option A (If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1) is correct because the equilibrium constant of an exothermic reaction decreases at high temperatures, and the provided equation correctly explains this relationship.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\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}

are the respective ionisation constants for the following reactions

\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}

The correct relationship between

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

\mathrm{pH}

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

solution is (For

\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}}.

In equilibrium condition pressure of

\text{O}_2

depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.