Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2026 · 23 January · Morning Shift · Chemistry · Chemical Equilibrium

Consider the general reaction given below at 400 K x ~A(~g)...

Consider the general reaction given below at 400 K
x A( g)⇌y B( g)x \mathrm{~A}(\mathrm{~g}) \rightleftharpoons y \mathrm{~B}(\mathrm{~g})
The values of Kp\mathrm{K}_{\mathrm{p}} and Kc\mathrm{K}_{\mathrm{c}} are studied under the same condition of temperature but variation in xx and yy.
(i) Kp=85.87\mathrm{K}_{\mathrm{p}}=85.87 and Kc=2.586\mathrm{K}_{\mathrm{c}}=2.586 appropriate units
(ii) Kp=0.862\mathrm{K}_{\mathrm{p}}=0.862 and Kc=28.62\mathrm{K}_{\mathrm{c}}=28.62 appropriate units.
The values of xx and yy in (i) and (ii) respectively are:

Options

  1. A((i)(ii)1,32,1\begin{array}{c} \text{(i)} & \text{(ii)} \\ 1,3 & 2,1 \end{array})
  2. B((i)(ii)4,14,1\begin{array}{c} \text{(i)} & \text{(ii)} \\ 4,1 & 4,1 \end{array})
  3. C((i)(ii)3,13,1\begin{array}{c} \text{(i)} & \text{(ii)} \\ 3,1 & 3,1 \end{array})
  4. D((i)(ii)1,22,1\begin{array}{c} \text{(i)} & \text{(ii)} \\ 1,2 & 2,1 \end{array})

Correct answer

D. ((i)(ii)1,22,1\begin{array}{c} \text{(i)} & \text{(ii)} \\ 1,2 & 2,1 \end{array})

Step-by-step solution

The relationship between KpK_p and KcK_c is given by Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}, where Δng=y−x\Delta n_g = y - x.
Given T=400T = 400 K and R=0.0821R = 0.0821 L atm K−1^{-1} mol−1^{-1}.
The value of RT=0.0821×400=32.84RT = 0.0821 \times 400 = 32.84.
For case (i): Kp=85.87K_p = 85.87 and Kc=2.586K_c = 2.586.
KpKc=85.872.586≈33.2\frac{K_p}{K_c} = \frac{85.87}{2.586} \approx 33.2.
Since 33.2≈32.84133.2 \approx 32.84^1, we have Δng=y−x=1\Delta n_g = y - x = 1.
Checking options for (i):
(1) y−x=3−1=2y-x = 3-1 = 2
(2) y−x=1−4=−3y-x = 1-4 = -3
(3) y−x=1−3=−2y-x = 1-3 = -2
(4) y−x=2−1=1y-x = 2-1 = 1. This matches.
For case (ii): Kp=0.862K_p = 0.862 and Kc=28.62K_c = 28.62.
KpKc=0.86228.62≈0.0301\frac{K_p}{K_c} = \frac{0.862}{28.62} \approx 0.0301.
Since 0.0301≈132.84=32.84−10.0301 \approx \frac{1}{32.84} = 32.84^{-1}, we have Δng=y−x=−1\Delta n_g = y - x = -1.
Checking option (4) for (ii): y−x=1−2=−1y-x = 1-2 = -1. This also matches.
Thus, for (i) x=1,y=2x=1, y=2 and for (ii) x=2,y=1x=2, y=1.
Practice Chemical Equilibrium on Quantrex Academy →

More from Chemical Equilibrium

Consider the following reactions in which all the reactants and products are present in… JEE Main 2026Solid carbon, CaO and CaCO _3 are mixed and allowed to attain equilibrium at T K.… JEE Main 2026In a closed flask at 600 K, one mole of X _2 Y _4 (g) attains equilibrium as given below… JEE Main 2026One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to… JEE Main 2026The reaction A(g) leftharpoons B(g) + C(g) was initiated with the amount ' a ' of A(g) .… JEE Main 2026The values of pressure equilibrium constant recorded at different temperatures for the… JEE Main 2026For the following reaction at 50° C and at 2 atm pressure, 2N_2O_5(g) leftharpoons… JEE Main 2026At T(K) , the equilibrium constant of A_2(g) + B_2(g) leftharpoons C(g) is 2.7 × 10^ -5 .… JEE Main 2026 Full Chemical Equilibrium list All JEE Main PYQs