The value of `K_(c)` for the reaction : `H_(2)(g)+I_(2)(g)hArr 2HI(g)` is 45.9 at 773 K. If one mole of `H_(2)`, two mole of `I_(2)` and three moles of HI are taken in a 1.0 L flask, the concentrations of HI at equilibrium at 773 K.

Options

A. 2.43 M
B. 4.36 M
C. 6.63 M
D. 7.82 M

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

### Step 2: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

### Step 3: Determine the initial concentrations
Given:
- Moles of \( H_2 = 1 \)
- Moles of \( I_2 = 2 \)
- Moles of \( HI = 3 \)
- Volume of the flask = 1 L

Since the volume is 1 L, the initial concentrations are:
- \([H_2] = 1 \, \text{M}\)
- \([I_2] = 2 \, \text{M}\)
- \([HI] = 3 \, \text{M}\)

### Step 4: Calculate the reaction quotient \( Q_c \)
The reaction quotient \( Q_c \) is calculated using the initial concentrations:
\[ Q_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(3)^2}{(1)(2)} = \frac{9}{2} = 4.5 \]

### Step 5: Compare \( Q_c \) and \( K_c \)
Given \( K_c = 45.9 \) and \( Q_c = 4.5 \):
Since \( K_c > Q_c \), the reaction will proceed in the forward direction to reach equilibrium.

### Step 6: Set up the equilibrium concentrations
Let \( x \) be the change in concentration as the reaction proceeds to equilibrium:
- At equilibrium, the concentrations will be:
- \([H_2] = 1 - x\)
- \([I_2] = 2 - x\)
- \([HI] = 3 + 2x\)

### Step 7: Write the equilibrium expression
Substituting the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{(3 + 2x)^2}{(1 - x)(2 - x)} \]
Setting this equal to the given \( K_c \):
\[ 45.9 = \frac{(3 + 2x)^2}{(1 - x)(2 - x)} \]

### Step 8: Solve for \( x \)
Cross-multiply and simplify:
\[ 45.9(1 - x)(2 - x) = (3 + 2x)^2 \]

Expanding both sides:
\[ 45.9(2 - 3x + x^2) = 9 + 12x + 4x^2 \]
\[ 91.8 - 137.7x + 45.9x^2 = 9 + 12x + 4x^2 \]
Rearranging gives:
\[ 41.9x^2 - 149.7x + 82.8 = 0 \]

### Step 9: Use the quadratic formula to solve for \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 41.9, b = -149.7, c = 82.8 \):
Calculate the discriminant:
\[ b^2 - 4ac = (-149.7)^2 - 4(41.9)(82.8) \]
\[ = 22410.09 - 13859.76 = 8549.33 \]

Now, calculate \( x \):
\[ x = \frac{149.7 \pm \sqrt{8549.33}}{2 \times 41.9} \]
\[ = \frac{149.7 \pm 92.5}{83.8} \]

Calculating the two possible values for \( x \):
1. \( x_1 = \frac{242.2}{83.8} \approx 2.89 \) (not possible since it exceeds initial moles)
2. \( x_2 = \frac{57.2}{83.8} \approx 0.68 \)

### Step 10: Calculate the equilibrium concentration of \( HI \)
Using \( x = 0.68 \):
\[ [HI]_{eq} = 3 + 2x = 3 + 2(0.68) = 3 + 1.36 = 4.36 \, \text{M} \]

### Final Answer
The concentration of \( HI \) at equilibrium is \( 4.36 \, \text{M} \).

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