For the equilibrium $\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{AB}_2(\mathrm{~g})$ the value of equilibrium constant at $300 \mathrm{k}$ is $3 \times 10^{-3} \mathrm{~mole}^{-2} \mathrm{~L}^2$ and value of equilibrium constant at $500 \mathrm{k}$ is $5 \times 10^{-5} \mathrm{~mol}^{-2} \mathrm{~L}^2$. What is the sign of enthalpy change for this reaction

Both (a) and (b) are possible at lower temperature it is positive and at high temperature it is negative.

JEE CHEMISTRY MCQ: Equilibrium

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For the equilibrium

\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{AB}_2(\mathrm{~g})

the value of equilibrium constant at

300 \mathrm{k}

3 \times 10^{-3} \mathrm{~mole}^{-2} \mathrm{~L}^2

and value of equilibrium constant at

500 \mathrm{k}

5 \times 10^{-5} \mathrm{~mol}^{-2} \mathrm{~L}^2

.

What is the sign of enthalpy change for this reaction

View answer and explanation

Correct option: B

Negative

The sign of the enthalpy change is negative because the equilibrium constant decreases as temperature increases, indicating an exothermic reaction.

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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