JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

(i) $X(g)\rightleftharpoons Y(g)+Z(g){K}_{p1}=3$

(ii) $A(g)\rightleftharpoons 2B(g){K}_{p2}=1$

If the degree of dissociation and initial concentration of both the reactants $X(g)$ and $A(g)$ are equal, then the ratio of the total pressure at equilibrium $(\frac{{p}_{1}}{{p}_{2}})$ is equal to $x:1$ . The value of $x$ is (Nearest integer)

Official previous-year question

Held on 1 Feb 2023 · Verified 6 Jul 2026.

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Solution

$X(g)\rightleftharpoons Y(g)+Z(g){K}_{{p}_{1}}=3$

$n$ Initial concentration

$n-\alpha n\alpha n\alpha n$ at equilibrium

Let total pressure be ${p}_{1}$

Using dalton's law of partial pressures and equation for equilibrium constant:

$\therefore {K}_{{p}_{1}}=\frac{{(\frac{\alpha }{1+\alpha }\times {p}_{1})}^{2}}{\frac{1-\alpha }{1+\alpha }{p}_{1}}$

$\Rightarrow 3=\frac{{\alpha }^{2}\times {p}_{1}}{1-{\alpha }^{2}}$ ....(X)

$A(g)\rightleftharpoons 2B(g){k}_{{p}_{2}}=1$

$n$ Initial mole

$x-\alpha n2\alpha n$ at equilibrium

Let total pressure be ${p}_{2}$

Using dalton's law of partial pressures and equation for equilibrium constant

$\therefore {k}_{{p}_{2}}=\frac{{(\frac{2\alpha }{1+\alpha }\times {p}_{2})}^{2}}{\frac{1-\alpha }{1+\alpha }\times {p}_{2}}$

$\Rightarrow 1=\frac{4{\alpha }^{2}\times {p}_{2}}{1-{\alpha }^{2}}$ ....(Y)

Thus, dividing equation (X) by equation(Y)

$\Rightarrow \frac{{k}_{{p}_{1}}}{{k}_{{p}_{2}}}=\frac{{p}_{1}}{4{p}_{2}}$

$\Rightarrow \frac{3}{1}=\frac{{p}_{1}}{4{p}_{2}}$

$\therefore \frac{{p}_{1}}{{p}_{2}}=12$

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Solution

$X(g)\rightleftharpoons Y(g)+Z(g){K}_{{p}_{1}}=3$

$n$ Initial concentration

$n-\alpha n\alpha n\alpha n$ at equilibrium

Let total pressure be ${p}_{1}$

Using dalton's law of partial pressures and equation for equilibrium constant:

$\therefore {K}_{{p}_{1}}=\frac{{(\frac{\alpha }{1+\alpha }\times {p}_{1})}^{2}}{\frac{1-\alpha }{1+\alpha }{p}_{1}}$

$\Rightarrow 3=\frac{{\alpha }^{2}\times {p}_{1}}{1-{\alpha }^{2}}$ ....(X)

$A(g)\rightleftharpoons 2B(g){k}_{{p}_{2}}=1$

$n$ Initial mole

$x-\alpha n2\alpha n$ at equilibrium

Let total pressure be ${p}_{2}$

Using dalton's law of partial pressures and equation for equilibrium constant

$\therefore {k}_{{p}_{2}}=\frac{{(\frac{2\alpha }{1+\alpha }\times {p}_{2})}^{2}}{\frac{1-\alpha }{1+\alpha }\times {p}_{2}}$

$\Rightarrow 1=\frac{4{\alpha }^{2}\times {p}_{2}}{1-{\alpha }^{2}}$ ....(Y)

Thus, dividing equation (X) by equation(Y)

$\Rightarrow \frac{{k}_{{p}_{1}}}{{k}_{{p}_{2}}}=\frac{{p}_{1}}{4{p}_{2}}$

$\Rightarrow \frac{3}{1}=\frac{{p}_{1}}{4{p}_{2}}$

$\therefore \frac{{p}_{1}}{{p}_{2}}=12$

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry