Equilibrium constant for the following equilibrium, $ 3 \mathrm{ClO}^{-}(a q) \rightleftharpoons 2 \mathrm{Cl}^{-}(a q)+\mathrm{ClO}_3^{-}(a q) $ is $10 \times 10^{28}$ at a temperature at which $R T=2500 \mathrm{~J} \mathrm{~mol}^{-1}$. Gibbs free energy change $\left(\Delta G^{\circ}\right)$ is

$-70.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$+30.395 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$+371.24 \mathrm{~kJ} \mathrm{~mol}^{-1}$

JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

Equilibrium constant for the following equilibrium,

3 \mathrm{ClO}^{-}(a q) \rightleftharpoons 2 \mathrm{Cl}^{-}(a q)+\mathrm{ClO}_3^{-}(a q)

10 \times 10^{28}

at a temperature at which

R T=2500 \mathrm{~J} \mathrm{~mol}^{-1}

.
Gibbs free energy change

\left(\Delta G^{\circ}\right)

View answer and explanation

Correct option: C

-161.21 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option C (-161.21 kJ mol⁻¹) is correct because it is calculated using the Gibbs free energy equation with the given equilibrium constant.

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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