To solve the problem, we need to find the concentration of HI at equilibrium for the reaction:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
Given:
- \( K_c = 45.9 \) at \( 773 \, K \)
- Initial moles: \( H_2 = 1 \, mol \), \( I_2 = 2 \, mol \), \( HI = 3 \, mol \)
- Volume of the flask = \( 1.0 \, L \)
### Step 1: Calculate Initial Concentrations
The initial concentrations of the reactants and products can be calculated as follows:
\[
\text{Concentration of } H_2 = \frac{1 \, mol}{1 \, L} = 1 \, M
\]
\[
\text{Concentration of } I_2 = \frac{2 \, mol}{1 \, L} = 2 \, M
\]
\[
\text{Concentration of } HI = \frac{3 \, mol}{1 \, L} = 3 \, M
\]
### Step 2: Set Up the Expression for \( K_c \)
The equilibrium expression for the reaction is given by:
\[
K_c = \frac{[HI]^2}{[H_2][I_2]}
\]
Substituting the initial concentrations into the equation:
\[
K_c = \frac{(3)^2}{(1)(2)} = \frac{9}{2} = 4.5
\]
### Step 3: Determine the Reaction Quotient \( Q \)
Since \( K_c \) (4.5) is less than the given \( K_c \) (45.9), the reaction will shift to the right (towards products).
### Step 4: Define Changes at Equilibrium
Let \( x \) be the amount of \( H_2 \) and \( I_2 \) that reacts at equilibrium. The changes in concentrations will be:
- \( [H_2] = 1 - x \)
- \( [I_2] = 2 - x \)
- \( [HI] = 3 + 2x \)
### Step 5: Write the Equilibrium Expression
Substituting these expressions into the equilibrium expression:
\[
K_c = \frac{(3 + 2x)^2}{(1 - x)(2 - x)} = 45.9
\]
### Step 6: Solve for \( x \)
Expanding and rearranging gives us:
\[
(3 + 2x)^2 = 45.9(1 - x)(2 - x)
\]
Expanding both sides:
\[
(3 + 2x)^2 = 9 + 12x + 4x^2
\]
\[
(1 - x)(2 - x) = 2 - 3x + x^2
\]
\[
45.9(2 - 3x + x^2) = 91.8 - 137.7x + 45.9x^2
\]
Setting the equation:
\[
9 + 12x + 4x^2 = 91.8 - 137.7x + 45.9x^2
\]
Rearranging gives:
\[
0 = 41.9x^2 - 149.7x + 82.8
\]
### Step 7: Use the Quadratic Formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where:
- \( a = 41.9 \)
- \( b = -149.7 \)
- \( c = 82.8 \)
Calculating the discriminant:
\[
D = b^2 - 4ac = (-149.7)^2 - 4(41.9)(82.8)
\]
Calculating \( D \):
\[
D = 22305.09 - 13859.76 = 8445.33
\]
Calculating \( x \):
\[
x = \frac{149.7 \pm \sqrt{8445.33}}{2 \times 41.9}
\]
Calculating the positive root:
\[
x \approx 0.68
\]
### Step 8: Calculate Equilibrium Concentration of HI
Substituting \( x \) back to find the concentration of HI:
\[
[HI]_{eq} = 3 + 2(0.68) = 3 + 1.36 = 4.36 \, M
\]
### Final Answer
The concentration of HI at equilibrium is approximately **4.36 M**.
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