At $600\,,\,{K_p}$ for the following reaction is 1 atm. $X\left( {\rm{g}} \right)\,\,\,\leftrightharpoons\,\,\,Y\left( {\rm{g}} \right) + Z\left( {\rm{g}} \right)$ At equilibrium, $50\,\% $ of $X\left( {\rm{g}} \right)$ is dissociated. The total pressure of the equilibrium system is $p$ atm. What is the partial pressure (in atm) of $X\left( {\rm{g}} \right)$ at equilibrium?

JEE CHEMISTRY MCQ: Equilibrium

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600\,,\,{K_p}

for the following reaction is 1 atm.

X\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,Y\left( \mathrm{g} \right) + Z\left( \mathrm{g} \right)

At equilibrium,

50\,\%

X\left( \mathrm{g} \right)

is dissociated. The total pressure of the equilibrium system is

p

atm. What is the partial pressure (in atm) of

X\left( \mathrm{g} \right)

at equilibrium?

View answer and explanation

Correct option: A

Option A (1) is correct because the partial pressure of X at equilibrium is calculated to be 1 atm using the given Kp and dissociation percentage.

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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