Question:
Observe the following reaction \(A(g) + B(g) \rightleftharpoons C(g)\). In a 1L closed flask, 2 moles of \(A(g)\) and 1 mole of \(B(g)\) were taken and heated to T(K). At equilibrium, the concentration of \(B(g)\) is equal to twice the equilibrium concentration of \(C(g)\). What is the value of \(K_c\)?
Observe the following reaction \(A(g) + B(g) \rightleftharpoons C(g)\). In a 1L closed flask, 2 moles of \(A(g)\) and 1 mole of \(B(g)\) were taken and heated to T(K). At equilibrium, the concentration of \(B(g)\) is equal to twice the equilibrium concentration of \(C(g)\). What is the value of \(K_c\)?
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Always use equilibrium concentrations in the \(K_c\) expression, not initial concentrations.
Updated On: Oct 7, 2026
- 0.3
- 0.6
- 1.2
- 1.5
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The Correct Option is A
Solution and Explanation
Concept:
The equilibrium constant \(K_c\) is determined by the ratio of the concentration of products to the concentration of reactants at equilibrium.
Step 1: Setup the ICE table.
Let \(x\) be the moles of \(C\) formed at equilibrium. Reaction: \(A + B \rightleftharpoons C\) Initial: \(2, 1, 0\) Change: \(-x, -x, +x\) Equilibrium: \(2-x, 1-x, x\)
Step 2: Apply the given equilibrium condition.
Given \([B] = 2[C]\): \[ 1 - x = 2x \Rightarrow 3x = 1 \Rightarrow x = 1/3 \] Equilibrium concentrations: \([A] = 2 - 1/3 = 5/3\) \([B] = 1 - 1/3 = 2/3\) \([C] = 1/3\)
Step 3: Calculate \(K_c\).
\[ K_c = \frac{[C]}{[A][B]} = \frac{1/3}{(5/3)(2/3)} = \frac{1/3}{10/9} \] \[ K_c = \frac{1}{3} \times \frac{9}{10} = \frac{3}{10} = 0.3 \] \[ \boxed{0.3} \]
Step 1: Setup the ICE table.
Let \(x\) be the moles of \(C\) formed at equilibrium. Reaction: \(A + B \rightleftharpoons C\) Initial: \(2, 1, 0\) Change: \(-x, -x, +x\) Equilibrium: \(2-x, 1-x, x\)
Step 2: Apply the given equilibrium condition.
Given \([B] = 2[C]\): \[ 1 - x = 2x \Rightarrow 3x = 1 \Rightarrow x = 1/3 \] Equilibrium concentrations: \([A] = 2 - 1/3 = 5/3\) \([B] = 1 - 1/3 = 2/3\) \([C] = 1/3\)
Step 3: Calculate \(K_c\).
\[ K_c = \frac{[C]}{[A][B]} = \frac{1/3}{(5/3)(2/3)} = \frac{1/3}{10/9} \] \[ K_c = \frac{1}{3} \times \frac{9}{10} = \frac{3}{10} = 0.3 \] \[ \boxed{0.3} \]
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