SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 53

####2. In which one of the following equilibria, $K_{p} \neq K_{c}$ ?

(a) $2 C (s) + O_{2} (g) ⇌ 2 C O (g)$

(2019 Main, 12 April II)

(b) $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$

(d) $2 N O (g) ⇌ N_{2} (g) + O_{2} (g)$

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Answer:

Correct Answer: 2. (b)

Solution:

Key Idea The relationship between $K_{p}$ and $K_{c}$ is

$K_{p} = K_{c} (R T)^{Δ n_{g}}$

where, $Δ n_{g} = n_{products} - n_{reactants}$

If $Δ n_{g} = 0$ then $K_{p} = K_{c}$

If $Δ n_{g} = +$ vethen $K_{p} > K_{c}$

If $Δ n_{g} = -$ vethen $K_{p} < K_{c}$

Consider the following equilibria reactions

(a) $2 C (s) + O_{2} (g) ⇌ 2 C O (g)$

$Δ n_{g} = n_{product} - n_{reactant} = 2 - (1) = 1$

$Δ n_{g} \neq 0 \Rightarrow S o, K_{p} \neq K_{c}$

(b) $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$

$Δ n_{g} = n_{product} - n_{reactant} = 2 - 2 = 0$

$Δ n_{g} = 0 \Rightarrow S o, K_{p} = K_{c}$

$Δ n_{g} = n_{product} - n_{reactant} = 2 - 2 = 0$

$Δ n_{g} = 0 \Rightarrow$ So, $K_{p} = K_{c}$

(d) $2 N O (g) ⇌ N_{2} (g) + O_{2} (g)$

$Δ n_{g} = n_{product} - n_{reactant} = 2 - 2 = 0$

$Δ n_{g} = 0 \Rightarrow$ So, $K_{p} = K_{c}$

Chemical and Ionic Equilibrium - Result Question 53

Answer:

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Chemical and Ionic Equilibrium - Result Question 53

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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