An equilibrium mixture for the reaction has one mole of hydrogen sulphide, 0.2 mole of H2 and 0.8 mole of S2 in a 2 litre vessel. The value of KC in mole litre-1 is

Chemical Equilibrium Problem: Hydrogen Sulfide DecompositionReaction:2H2S(g) ⇌ 2H2(g) + S2(g)Given: Equilibrium mixture in a 2 L vessel contains:1 mol of H2S0.2 mol of H20.8 mol of S2Find: The value of KC in mol·L-1Step 1: Calculate Equilibrium ConcentrationsUsing the formula: Concentration = Number of moles / Volume (L)SpeciesMolesVolume (L)Concentration (mol/L)[H2S]120.5[H2]0.220.1[S2]0.820.4Step 2: Write Equilibrium Constant ExpressionFor the reaction: 2H2S(g) ⇌ 2H2(g) + S2(g)KC = [H2]2 × [S2] / [H2S]2Step 3: Substitute Values and CalculateKC = (0.1)2 × (0.4) / (0.5)2KC = (0.01) × (0.4) / (0.25)KC = 0.004 / 0.25 = 0.016Final Answer:The value of KC is 0.016 mol·L-1Note: This low value indicates that at equilibrium, the reactant (H2S) is favored over the products, meaning the decomposition is limited under these conditions.

An equilibrium mixture for the reaction has one mole of hydrogen sulphide, 0.2 mole of H2 and 0.8 mole of S2 in a 2 litre vessel. The value of KC in mole litre-1 is

Chemical Equilibrium Problem: Hydrogen Sulfide DecompositionReaction:2H2S(g) ⇌ 2H2(g) + S2(g)Given: Equilibrium mixture in a 2 L vessel contains:1 mol of H2S0.2 mol of H20.8 mol of S2Find: The value of KC in mol·L-1Step 1: Calculate Equilibrium ConcentrationsUsing the formula: Concentration = Number of moles / Volume (L)SpeciesMolesVolume (L)Concentration (mol/L)[H2S]120.5[H2]0.220.1[S2]0.820.4Step 2: Write Equilibrium Constant ExpressionFor the reaction: 2H2S(g) ⇌ 2H2(g) + S2(g)KC = [H2]2 × [S2] / [H2S]2Step 3: Substitute Values and CalculateKC = (0.1)2 × (0.4) / (0.5)2KC = (0.01) × (0.4) / (0.25)KC = 0.004 / 0.25 = 0.016Final Answer:The value of KC is 0.016 mol·L-1Note: This low value indicates that at equilibrium, the reactant (H2S) is favored over the products, meaning the decomposition is limited under these conditions.

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An equilibrium mixture for the reaction $\large 2{H_2}{S_{(g)}}\, \rightleftharpoons \,2{H_{2(g)}}\, + \,{S_{2(g)}}$ has one mole of hydrogen sulphide, 0.2 mole of H₂ and 0.8 mole of S₂ in a 2 litre vessel. The value of K_C in mole litre^-1 is

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0.004

0.160

0.016

0.080

answer is B.

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Detailed Solution

Chemical Equilibrium Problem: Hydrogen Sulfide Decomposition

Reaction:

2H₂S_(g) ⇌ 2H_2(g) + S_2(g)

Given: Equilibrium mixture in a 2 L vessel contains:

1 mol of H₂S
0.2 mol of H₂
0.8 mol of S₂

Find: The value of K_C in mol·L^-1

Step 1: Calculate Equilibrium Concentrations

Using the formula: Concentration = Number of moles / Volume (L)

Species	Moles	Volume (L)	Concentration (mol/L)
[H₂S]	1	2	0.5
[H₂]	0.2	2	0.1
[S₂]	0.8	2	0.4

Step 2: Write Equilibrium Constant Expression

For the reaction: 2H₂S_(g) ⇌ 2H_2(g) + S_2(g)

K_C = [H₂]² × [S₂] / [H₂S]²

Step 3: Substitute Values and Calculate

K_C = (0.1)² × (0.4) / (0.5)²

K_C = (0.01) × (0.4) / (0.25)

K_C = 0.004 / 0.25 = 0.016

Final Answer:The value of K_C is 0.016 mol·L^-1Note: This low value indicates that at equilibrium, the reactant (H₂S) is favored over the products, meaning the decomposition is limited under these conditions.

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