JEE Main 2026 · 5 Apr · Shift 2 · Chemistry Q56

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ChemistrySingle correct+4 / -1Question 56 of 75

The reaction $\ce{A(g) <=> B(g) + C(g)}$ 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 Kp of the reaction can be calculated from the expression :

\dfrac{x^2}{a^2+x^2}\times p

\dfrac{x^2}{a^2-x^2}\times p

\dfrac{a+x^2}{x^2}\times p

\dfrac{a^2-x^2}{x^2}\times p

Answer and worked solution

Official answer: B

Equilibrium amounts. A: $a-x$ ; B: x; C: x. Total: $a+x$ .

Kp.
$K_p=\frac{\left(\frac x{a+x}p\right)^2}{\frac{a-x}{a+x}p}=\frac{x^2}{(a+x)(a-x)}p=\frac{x^2}{a^2-x^2}\,p.$

(B)

Topics:Law of mass action and Kc, Kp

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JEE Main 2026 · 5 Apr · Shift 2, Chemistry question 56