One mole of `H_2.` two moles of `I_2` and three moles of Hl injected in a one litre flask . What will be the concentrations of `H_2, I_2` and Hl at equilibrium at ` 490 ^(@) C .` The equilibrium constant for the reaction at ` 490 ^(@) C ` is 45. 9

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Set up the initial concentrations
We start with:
- 1 mole of \( H_2 \)
- 2 moles of \( I_2 \)
- 3 moles of \( HI \)

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

### Step 3: Define changes in concentration at equilibrium
Let \( x \) be the amount of \( H_2 \) and \( I_2 \) that reacts at equilibrium. Thus:
- The change in concentration for \( H_2 \) and \( I_2 \) will be \( -x \).
- The change in concentration for \( HI \) will be \( +2x \).

At equilibrium, the concentrations will be:
- \([H_2] = 1 - x\)
- \([I_2] = 2 - x\)
- \([HI] = 3 + 2x\)

### Step 4: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{(3 + 2x)^2}{(1 - x)(2 - x)} = 45.9 \]

### Step 5: Expand and rearrange the equation
Expanding the equation:
\[ (3 + 2x)^2 = 9 + 12x + 4x^2 \]
And the denominator:
\[ (1 - x)(2 - x) = 2 - 3x + x^2 \]
Thus, we have:
\[ \frac{9 + 12x + 4x^2}{2 - 3x + x^2} = 45.9 \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 9 + 12x + 4x^2 = 45.9(2 - 3x + x^2) \]
Expanding the right side:
\[ 9 + 12x + 4x^2 = 91.8 - 137.7x + 45.9x^2 \]
Rearranging gives:
\[ (4x^2 - 45.9x^2) + (12x + 137.7x) + (9 - 91.8) = 0 \]
This simplifies to:
\[ -41.9x^2 + 149.7x - 82.8 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = -41.9 \), \( b = 149.7 \), and \( c = -82.8 \):
\[ x = \frac{-149.7 \pm \sqrt{(149.7)^2 - 4 \cdot (-41.9) \cdot (-82.8)}}{2 \cdot (-41.9)} \]
Calculating the discriminant:
\[ x = \frac{-149.7 \pm \sqrt{22415.09 - 13859.76}}{-83.8} \]
\[ x = \frac{-149.7 \pm \sqrt{8545.33}}{-83.8} \]
Calculating the square root and solving for \( x \):
\[ x \approx 0.684 \, \text{(discarding the negative root)} \]

### Step 8: Calculate equilibrium concentrations
Now we substitute \( x \) back to find the equilibrium concentrations:
- \([H_2] = 1 - x = 1 - 0.684 = 0.316 \, \text{mol/L}\)
- \([I_2] = 2 - x = 2 - 0.684 = 1.316 \, \text{mol/L}\)
- \([HI] = 3 + 2x = 3 + 2(0.684) = 4.368 \, \text{mol/L}\)

### Final Answer:
- \([H_2] = 0.316 \, \text{mol/L}\)
- \([I_2] = 1.316 \, \text{mol/L}\)
- \([HI] = 4.368 \, \text{mol/L}\)