Prove that the pressure necessary to obtain $50 \%$ dissociation of $\mathrm{PCl}_5$ at $250^{\circ} \mathrm{C}$ is numerically three times of $K_p$.

$\mathrm{P}=4 \mathrm{~K}_{\mathrm{P}}$

$\mathrm{P}=5 \mathrm{~K}_{\mathrm{P}}$

$\mathrm{P}=6 \mathrm{~K}_{\mathrm{P}}$

Back to Equilibrium

JEE CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

Prove that the pressure necessary to obtain

50 \%

dissociation of

\mathrm{PCl}_5

250^{\circ} \mathrm{C}

is numerically three times of

K_p

View answer and explanation

Correct option: A

\mathrm{P}=3 \mathrm{~K}_{\mathrm{P}}

Option A (P = 3Kp) is correct because the equilibrium constant Kp is related to the total pressure P by the equation Kp = P/3 when the degree of dissociation is 50%.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

(i)

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

(ii)

Then,

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\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}

The correct relationship between

K{a_1},\,\,K{a_2}\,,K{a_3}

is The

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\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

) Reaction

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\text{O}_2

depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.