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?

• A. \(0.04 \)
• B. \(0.4 \)
• C. \(0.02 \)
• D. \(0.2 \)

Hint:

When performing equilibrium calculations, remember that \(K_c\) is calculated using molar concentrations (\(mol/L\)), not just the number of moles.

Answer 1

The Correct Option is A

Step 1: Write the equilibrium constant.
For $2SO_2+O_2\rightleftharpoons 2SO_3$, the constant is $K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}$. Concentration is moles divided by the 10 L volume.
Step 2: Name the unknowns.
Let moles of $SO_2$ be $n$. The problem says $SO_3$ is twice that, so $SO_3=2n$. Let $O_2$ moles be $x$.
Step 3: Put concentrations in.
\[ 100=\frac{(2n/10)^2}{(n/10)^2\,(x/10)} \]
Step 4: Cancel the n terms.
The $(2n/10)^2$ over $(n/10)^2$ leaves $4$. So \[ 100=\frac{4}{x/10} \]
Step 5: Simplify the volume part.
\[ 100=\frac{4\times10}{x}=\frac{40}{x} \]
Step 6: Solve for x.
\[ x=\frac{40}{100}=0.4\ \text{mol} \] \[ \boxed{0.4\ \text{mol of }O_2} \]