JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

The reaction $A(g) \rightleftharpoons B(g) + C(g)$ was initiated with the amount ' $a$ ' of $A(g)$ . At equilibrium it is found that the amount of $A(g)$ remaining is $(a - x)$ at a total pressure of $p$ .

The equilibrium constant $K_p$ of the reaction can be calculated from the expression :

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{x^2}{a^2 + x^2} \times p$
B
$\dfrac{x^2}{a^2 - x^2} \times p$
C
$\dfrac{a + x^2}{x^2} \times p$
D
$\dfrac{a^2 - x^2}{x^2} \times p$

Did you get this right?

Solution

The reaction is $A(g) \rightleftharpoons B(g) + C(g)$

Initial moles: $a$ for $A$ , $0$ for $B$ , $0$ for $C$

Moles at equilibrium: $(a - x)$ for $A$ , $x$ for $B$ , $x$ for $C$

Total moles at equilibrium $= (a - x) + x + x = a + x$

Partial pressures at equilibrium:

$p_A = \dfrac{a - x}{a + x} p$

$p_B = \dfrac{x}{a + x} p$

$p_C = \dfrac{x}{a + x} p$

The equilibrium constant $K_p$ is given by:

$K_p = \dfrac{p_B \cdot p_C}{p_A}$

$K_p = \dfrac{\left(\dfrac{x}{a + x} p\right) \left(\dfrac{x}{a + x} p\right)}{\dfrac{a - x}{a + x} p}$

$K_p = \dfrac{x^2}{(a + x)(a - x)} p$

$K_p = \dfrac{x^2}{a^2 - x^2} p$

Answer: $\dfrac{x^2}{a^2 - x^2} \times p$

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

The equilibrium constant $K_p$ of the reaction can be calculated from the expression :

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{x^2}{a^2 + x^2} \times p$
B
$\dfrac{x^2}{a^2 - x^2} \times p$
C
$\dfrac{a + x^2}{x^2} \times p$
D
$\dfrac{a^2 - x^2}{x^2} \times p$

Did you get this right?

Solution

The reaction is $A(g) \rightleftharpoons B(g) + C(g)$

Initial moles: $a$ for $A$ , $0$ for $B$ , $0$ for $C$

Moles at equilibrium: $(a - x)$ for $A$ , $x$ for $B$ , $x$ for $C$

Total moles at equilibrium $= (a - x) + x + x = a + x$

Partial pressures at equilibrium:

$p_A = \dfrac{a - x}{a + x} p$

$p_B = \dfrac{x}{a + x} p$

$p_C = \dfrac{x}{a + x} p$

The equilibrium constant $K_p$ is given by:

$K_p = \dfrac{p_B \cdot p_C}{p_A}$

$K_p = \dfrac{\left(\dfrac{x}{a + x} p\right) \left(\dfrac{x}{a + x} p\right)}{\dfrac{a - x}{a + x} p}$

$K_p = \dfrac{x^2}{(a + x)(a - x)} p$

$K_p = \dfrac{x^2}{a^2 - x^2} p$

Answer: $\dfrac{x^2}{a^2 - x^2} \times p$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

Options

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

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