`1` mol of `H_(2), 2` mol of `I_(2)` and `3` mol of HI were taken in a `1-L` flask. If the value of `K_(c)` for the equation `H_(2)(g)+I_(2)(g) hArr 2HI(g)` is `50` at `440^(@)C`, what will be the concentration of each specie at equilibrium?

Explanation

To solve the problem step by step, we will follow the systematic approach of using the equilibrium constant expression and the initial conditions given in the problem.

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

### Step 2: Identify initial concentrations
We are given:
- 1 mole of \( H_2 \)
- 2 moles of \( I_2 \)
- 3 moles of \( HI \)

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

### Step 3: Set up the change in concentration at equilibrium
Let \( x \) be the amount of \( H_2 \) and \( I_2 \) that reacts at equilibrium. The changes in concentration will be:
- For \( H_2 \): \( [H_2] = 1 - x \)
- For \( I_2 \): \( [I_2] = 2 - x \)
- For \( HI \): \( [HI] = 3 + 2x \)

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Given that \( K_c = 50 \), we can substitute the equilibrium concentrations into the expression:
\[ 50 = \frac{(3 + 2x)^2}{(1 - x)(2 - x)} \]

### Step 5: Solve for \( x \)
Now we will solve the equation:
\[ 50 = \frac{(3 + 2x)^2}{(1 - x)(2 - x)} \]
Cross-multiplying gives:
\[ 50(1 - x)(2 - x) = (3 + 2x)^2 \]
Expanding both sides:
\[ 50(2 - 3x + x^2) = 9 + 12x + 4x^2 \]
\[ 100 - 150x + 50x^2 = 9 + 12x + 4x^2 \]
Rearranging gives:
\[ 46x^2 - 162x + 91 = 0 \]

### Step 6: Use the quadratic formula to find \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 46 \)
- \( b = -162 \)
- \( c = 91 \)

Calculating the discriminant:
\[ b^2 - 4ac = (-162)^2 - 4(46)(91) = 26244 - 16736 = 9508 \]
Now applying the quadratic formula:
\[ x = \frac{162 \pm \sqrt{9508}}{92} \]
Calculating \( \sqrt{9508} \approx 97.5 \):
\[ x = \frac{162 \pm 97.5}{92} \]
Calculating the two possible values for \( x \):
1. \( x_1 = \frac{259.5}{92} \approx 2.82 \) (not possible since it exceeds initial moles)
2. \( x_2 = \frac{64.5}{92} \approx 0.7 \)

Thus, \( x \approx 0.7 \).

### Step 7: Calculate equilibrium concentrations
Now substituting \( x \) back into the expressions for equilibrium concentrations:
- \([H_2] = 1 - x = 1 - 0.7 = 0.3 \, \text{mol/L}\)
- \([I_2] = 2 - x = 2 - 0.7 = 1.3 \, \text{mol/L}\)
- \([HI] = 3 + 2x = 3 + 2(0.7) = 4.4 \, \text{mol/L}\)

### Final Answer
At equilibrium, the concentrations are:
- \([H_2] = 0.3 \, \text{mol/L}\)
- \([I_2] = 1.3 \, \text{mol/L}\)
- \([HI] = 4.4 \, \text{mol/L}\)

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