Consider the general reaction given below at 400 K xA(g) leftharpoons yB(g) The values of Kp and Kc are studied under the same condition of temperature but variation in x and y. (i) Kp = 85.87 and Kc

For the reaction xA(g) leftharpoons yB(g), Delta ng = y - x. We use R = 0.0821 L atm K-1 mol-1 and T = 400 K. RT = 0.0821 x 400 = 32.84. In case (i), Kp = 85.87 and Kc = 2.586. Since Kp > Kc, Delta ng > 0, implying y - x > 0. Solving gives y - x = 1. Looking at options for (i), "1, 2" (where x=1, y=2) fits perfectly because 2 - 1 = 1. In case (ii), Kp = 0.862 and Kc = 28.62. Since Kp Kc y > x. Kp < Kc y < x. Only option (4) satisfies both conditions simultaneously without even needing full calcu

Consider the general reaction given below at 400 K xA(g) leftharpoons yB(g) The values of Kp and Kc are studied under the same condition of temperature but variation in x and y. (i) Kp = 85.87 and Kc

For the reaction xA(g) leftharpoons yB(g), Delta ng = y - x. We use R = 0.0821 L atm K-1 mol-1 and T = 400 K. RT = 0.0821 x 400 = 32.84. In case (i), Kp = 85.87 and Kc = 2.586. Since Kp > Kc, Delta ng > 0, implying y - x > 0. Solving gives y - x = 1. Looking at options for (i), "1, 2" (where x=1, y=2) fits perfectly because 2 - 1 = 1. In case (ii), Kp = 0.862 and Kc = 28.62. Since Kp Kc y > x. Kp < Kc y < x. Only option (4) satisfies both conditions simultaneously without even needing full calcu

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Equilibrium appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Relation between Kp and Kc.

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Consider the general reaction given below at 400 K $xA(g) leftharpoons yB(g)$ $xA(g) \rightleftharpoons yB(g)$ The values of $Kₚ$ $K_{p}$ and $Kc$ $K_{c}$ are studied under the same condition of temperature but variation in x and y. (i) $Kₚ = 85.87$ $K_{p} = 85.87$ and $Kc = 2.586$ $K_{c} = 2.586$ appropriate units (ii) $Kₚ = 0.862$ $K_{p} = 0.862$ and $Kc = 28.62$ $K_{c} = 28.62$ appropriate units The value of x and y in (i) and (ii) respectively are:

A. $(i) 3, 1 ; (ii) 3, 1$ $\text{(i) 3, 1 ; (ii) 3, 1}$
B. $(i) 4, 1 ; (ii) 4, 1$ $\text{(i) 4, 1 ; (ii) 4, 1}$
C. $(i) 1, 3 ; (ii) 2, 1$ $\text{(i) 1, 3 ; (ii) 2, 1}$
D. $(i) 1, 2 ; (ii) 2, 1$ $\text{(i) 1, 2 ; (ii) 2, 1}$

Solution & Explanation

Related Formula

Kₚ = Kc (RT)Δ ng

Core Logic

For the reaction $xA(g) leftharpoons yB(g)$ $xA(g) \rightleftharpoons yB(g)$, $Δ ng = y - x$ $\Delta n_g = y - x$. We use $R = 0.0821 L atm K⁻¹ mol⁻¹$ and $T = 400 K$ $T = 400 \text{ K}$.

$RT = 0.0821 \times 400 = 32.84$ $RT = 0.0821 \times 400 = 32.84$.

Step 1: Case (i) Analysis

In case (i), $Kₚ = 85.87$ $K_p = 85.87$ and $Kc = 2.586$ $K_c = 2.586$. Since $Kₚ > Kc$ $K_p > K_c$, $Δ ng > 0$ $\Delta n_g > 0$, implying $y - x > 0$ $y - x > 0$.

85.87 = 2.586 \times (32.84)y-x

33.20 \approx (32.84)y-x

Solving gives $y - x = 1$ $y - x = 1$. Looking at options for (i), "1, 2" (where $x=1, y=2$ $x=1, y=2$) fits perfectly because $2 - 1 = 1$ $2 - 1 = 1$.

Step 2: Case (ii) Analysis

In case (ii), $Kₚ = 0.862$ $K_p = 0.862$ and $Kc = 28.62$ $K_c = 28.62$. Since $Kₚ < Kc$ $K_p < K_c$, $Δ ng < 0$ $\Delta n_g < 0$, implying $y - x < 0$ $y - x < 0$. Looking at options for (ii), "2, 1" (where $x=2, y=1$ $x=2, y=1$) gives $y - x = -1$ $y - x = -1$.

Kₚ = Kc \times (32.84)⁻¹ 0.862 = 28.62 / 32.84 \approx 0.87

This matches correctly.

Pattern Recognition

Just check the signs! $Kₚ > Kc \Rightarrow y > x$ $K_p > K_c \Rightarrow y > x$. $Kₚ < Kc \Rightarrow y < x$ $K_p < K_c \Rightarrow y < x$. Only option (4) satisfies both conditions simultaneously without even needing full calculation.

Chapter Mix

Class 11 Chemistry: Equilibrium

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Equilibrium

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Step 2: Case (ii) Analysis

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