To find the equilibrium constant \( K_c \) for the reaction
\[
\text{Fe}^{3+} + \text{SCN}^- \rightleftharpoons \text{FeSCN}^{2+}
\]
we start with the given information:
- Initial moles of \( \text{Fe}^{3+} = 3.1 \, \text{mol} \)
- Initial moles of \( \text{SCN}^- = 3.2 \, \text{mol} \)
- At equilibrium, moles of \( \text{FeSCN}^{2+} = 3 \, \text{mol} \)
### Step 1: Set up the initial concentrations
Since the total volume of the solution is 1 L, the initial concentrations are:
\[
[\text{Fe}^{3+}]_0 = 3.1 \, \text{mol/L} = 3.1 \, \text{M}
\]
\[
[\text{SCN}^-]_0 = 3.2 \, \text{mol/L} = 3.2 \, \text{M}
\]
\[
[\text{FeSCN}^{2+}]_0 = 0 \, \text{M}
\]
### Step 2: Set up the change in concentrations
Let \( x \) be the amount of \( \text{FeSCN}^{2+} \) formed at equilibrium. From the problem, we know \( x = 3 \, \text{mol} \).
The changes in concentrations will be:
- For \( \text{Fe}^{3+} \): \( 3.1 - x = 3.1 - 3 = 0.1 \, \text{M} \)
- For \( \text{SCN}^- \): \( 3.2 - x = 3.2 - 3 = 0.2 \, \text{M} \)
- For \( \text{FeSCN}^{2+} \): \( 0 + x = 3 \, \text{M} \)
### Step 3: Write the equilibrium concentrations
At equilibrium, we have:
\[
[\text{Fe}^{3+}] = 0.1 \, \text{M}
\]
\[
[\text{SCN}^-] = 0.2 \, \text{M}
\]
\[
[\text{FeSCN}^{2+}] = 3 \, \text{M}
\]
### Step 4: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:
\[
K_c = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]}
\]
### Step 5: Substitute the equilibrium concentrations into the expression
Now, substituting the equilibrium concentrations into the equation:
\[
K_c = \frac{3}{(0.1)(0.2)}
\]
### Step 6: Calculate \( K_c \)
Calculating the denominator:
\[
(0.1)(0.2) = 0.02
\]
Now substituting back into the equation for \( K_c \):
\[
K_c = \frac{3}{0.02} = 150
\]
### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[
K_c = 150
\]
