Question:
At \(T(K)\), in a 10 L flask, the following equilibrium is established:
\(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)\).
The value of \(K_{c}\) for this reaction is 100. At equilibrium, the number of moles of \(SO_{3}(g)\) is equal to twice the number of moles of \(SO_{2}(g)\). What is the number of moles of \(O_{2}(g)\) at equilibrium?
At \(T(K)\), in a 10 L flask, the following equilibrium is established:
\(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)\).
The value of \(K_{c}\) for this reaction is 100. At equilibrium, the number of moles of \(SO_{3}(g)\) is equal to twice the number of moles of \(SO_{2}(g)\). What is the number of moles of \(O_{2}(g)\) at equilibrium?
\(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)\).
The value of \(K_{c}\) for this reaction is 100. At equilibrium, the number of moles of \(SO_{3}(g)\) is equal to twice the number of moles of \(SO_{2}(g)\). What is the number of moles of \(O_{2}(g)\) at equilibrium?
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When performing equilibrium calculations, remember that \(K_c\) is calculated using molar concentrations (\(mol/L\)), not just the number of moles.
Updated On: Jun 8, 2026
- \(0.04 \)
- \(0.4 \)
- \(0.02 \)
- \(0.2 \)
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The Correct Option is A
Solution and Explanation
Concept:
The equilibrium constant (\(K_c\)) is defined by the ratio of the molar concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. Since the volume is 10 L, the concentration \([X] = n_X / 10\).
Step 1: Express the equilibrium concentrations.
Let the equilibrium moles of \(SO_2\) be \(n\). Then, the moles of \(SO_3\) will be \(2n\). Let the moles of \(O_2\) at equilibrium be \(x\).
Step 2: Apply the \(K_c\) expression.
\[ K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} \Rightarrow 100 = \frac{(2n/10)^2}{(n/10)^2 \cdot (x/10)} \]
Step 3: Solve for \(x\).
Simplifying the expression: \[ 100 = \frac{4n^2/100}{(n^2/100) \cdot (x/10)} \Rightarrow 100 = \frac{4}{x/10} \Rightarrow 100 = \frac{40}{x} \] \[ x = \frac{40}{100} = 0.4 \text{ moles of } O_2 \text{ in the flask.} \]
Step 1: Express the equilibrium concentrations.
Let the equilibrium moles of \(SO_2\) be \(n\). Then, the moles of \(SO_3\) will be \(2n\). Let the moles of \(O_2\) at equilibrium be \(x\).
Step 2: Apply the \(K_c\) expression.
\[ K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} \Rightarrow 100 = \frac{(2n/10)^2}{(n/10)^2 \cdot (x/10)} \]
Step 3: Solve for \(x\).
Simplifying the expression: \[ 100 = \frac{4n^2/100}{(n^2/100) \cdot (x/10)} \Rightarrow 100 = \frac{4}{x/10} \Rightarrow 100 = \frac{40}{x} \] \[ x = \frac{40}{100} = 0.4 \text{ moles of } O_2 \text{ in the flask.} \]
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