JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

The dissociation constant of

\mathrm{NH}_3

27^{\circ} \mathrm{C}

from the following data:

\begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \rightleftharpoons \mathrm{NH}_4^{+} ; \Delta H^{\circ}=-52.21 \mathrm{~kJ} / \mathrm{mol} ; \Delta S^{\circ}=+1.6 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1} \\ & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} ; \Delta H^{\circ}=54.70 \mathrm{~kJ} / \mathrm{mol} ; \Delta S^{\circ}=-76.3 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}

Given:

\mathrm{R}=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol}

View answer and explanation

Correct option: B

\mathrm{e}^{-10}

Option B (e^-10) is correct because it is the calculated dissociation constant of NH3 based on the given thermodynamic data.

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