JEE CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

Find the value of

\mathrm{K}_{\mathrm{p}}

for dissociation of:

2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2

,

if the equilibrium concentration of

\mathrm{HI}, \mathrm{H}_2

and

\mathrm{I}_2

are

3.53,0.4789

and

0.4789 \mathrm{~mole ~litre}^{-1}

respectively.

View answer and explanation

Correct option: A

1.814 \times 10^{-2}

Option A (

1.814 \times 10^{-2}

) is correct because Kp is equal to Kc when delta n is zero, and Kc is calculated from the equilibrium concentrations.

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