JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

K_p

has the value of

10^{-6} {~atm}^3

and

10^{-4} {~atm}^3

298 {~K}

and

323 {~K}

respectively for the reaction

{CuSO}_4 \cdot 3 {H}_2 {O}({s}) \rightleftharpoons {CuSO}_4({~s})+3 {H}_2 {O}({g})

\Delta_{{r}} H^{\circ}

for the reaction is :

View answer and explanation

Correct option: C

147.41 {~kJ} / {mol}

Option C (

147.41 {~kJ} / {mol}

) is correct because it is calculated using the Van't Hoff equation relating the change in Kp with temperature to the enthalpy change.

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