The equilibrium ${\rm{PC}}{{\rm{l}}_5}\left( {\rm{g}} \right)\;\;\leftrightharpoons{\rm{PC}}{{\rm{l}}_3}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)$ shows that ${K_p}$ is $24.63$ times the value of ${K_c}$ at a particular temperature $T$. Then $T$ (in $K$) is:

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Application +4 | -1

The equilibrium

\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\;\;\leftrightharpoons\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)

shows that

{K_p}

24.63

times the value of

{K_c}

at a particular temperature

T

. Then

T

(in

K

) is:

View answer and explanation

Correct option: C

300

Option A (

80 \%

) is correct because it's the degree of association calculated using the van't Hoff factor and the freezing point depression.

Practice All Chapter MCQs →

\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