The solids dissociate as follows: $A\left( s \right)\,\leftrightharpoons\,B\left( g \right) + C\left( g \right);\,\,{K_{{P_1}}} = x\,\,\,{\rm{at}}{{\rm{m}}^2}$ $D\left( s \right)\,\leftrightharpoons\,C\left( g \right) + E\left( g \right);\,\,{K_{{P_2}}} = y\,\,\,{\rm{at}}{{\rm{m}}^2}$ The total pressure when both the solids dissociate simultaneously is

$2\left( {\sqrt {x + y} } \right){\rm{ \,atm}}$

$\left( {x + y} \right){\rm{ \,atm}}$

JEE CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

The solids dissociate as follows:

A\left( s \right)\,\leftrightharpoons\,B\left( g \right) + C\left( g \right);\,\,{K_{{P_1}}} = x\,\,\,\mathrm{at}{\mathrm{m}^2}

D\left( s \right)\,\leftrightharpoons\,C\left( g \right) + E\left( g \right);\,\,{K_{{P_2}}} = y\,\,\,\mathrm{at}{\mathrm{m}^2}

The total pressure when both the solids dissociate simultaneously is

View answer and explanation

Correct option: D

2\left( {\sqrt {x + y} } \right)\mathrm{ \,atm}

Option D (

2(\sqrt {x + y} )\mathrm{ atm}

) is correct because it represents the total pressure derived from the equilibrium constants of the two dissociation reactions.

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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