`2ICl rarr I_(2) + Cl_(2)` `K_(C) = 0.14`
Initial concentration of ICl is `0.6` M then equilibrium concentration of `I_(2)` is:

Options

A. 0.37M
B. 0.128M
C. 0.224M
D. 0.748M

Explanation

To solve the equilibrium problem for the reaction \(2 \text{ICl} \rightleftharpoons \text{I}_2 + \text{Cl}_2\) with a given equilibrium constant \(K_c = 0.14\) and an initial concentration of \(\text{ICl} = 0.6 \, \text{M}\), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \(K_c\).
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2} \]

### Step 2: Set up the initial concentrations and changes at equilibrium.
- Initial concentration of \(\text{ICl} = 0.6 \, \text{M}\)
- Initial concentrations of \(\text{I}_2\) and \(\text{Cl}_2 = 0 \, \text{M}\)

Let \(x\) be the change in concentration of \(\text{I}_2\) and \(\text{Cl}_2\) at equilibrium. Since 2 moles of \(\text{ICl}\) produce 1 mole of \(\text{I}_2\) and 1 mole of \(\text{Cl}_2\), we can write:
- At equilibrium:
- \([\text{ICl}] = 0.6 - 2x\)
- \([\text{I}_2] = x\)
- \([\text{Cl}_2] = x\)

### Step 3: Substitute these expressions into the \(K_c\) expression.
Substituting the equilibrium concentrations into the \(K_c\) expression gives:
\[ 0.14 = \frac{x \cdot x}{(0.6 - 2x)^2} \]
This simplifies to:
\[ 0.14 = \frac{x^2}{(0.6 - 2x)^2} \]

### Step 4: Cross-multiply and simplify.
Cross-multiplying gives:
\[ 0.14(0.6 - 2x)^2 = x^2 \]
Expanding the left side:
\[ 0.14(0.36 - 2.4x + 4x^2) = x^2 \]
This leads to:
\[ 0.0504 - 0.336x + 0.56x^2 = x^2 \]
Rearranging gives:
\[ 0.56x^2 - x^2 - 0.336x + 0.0504 = 0 \]
Which simplifies to:
\[ -0.44x^2 - 0.336x + 0.0504 = 0 \]
Multiplying through by -1 gives:
\[ 0.44x^2 + 0.336x - 0.0504 = 0 \]

### Step 5: Use the quadratic formula to solve for \(x\).
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 0.44\), \(b = 0.336\), and \(c = -0.0504\):
\[ x = \frac{-0.336 \pm \sqrt{(0.336)^2 - 4(0.44)(-0.0504)}}{2(0.44)} \]
Calculating the discriminant:
\[ (0.336)^2 - 4(0.44)(-0.0504) = 0.112896 + 0.088704 = 0.2016 \]
Now calculating \(x\):
\[ x = \frac{-0.336 \pm \sqrt{0.2016}}{0.88} \]
Calculating \(\sqrt{0.2016} \approx 0.448\):
\[ x = \frac{-0.336 \pm 0.448}{0.88} \]
Calculating the two possible values for \(x\):
1. \(x = \frac{0.112}{0.88} \approx 0.127\)
2. \(x = \frac{-0.784}{0.88}\) (not valid since concentration cannot be negative)

Thus, we find:
\[ x \approx 0.128 \, \text{M} \]

### Step 6: Conclusion
The equilibrium concentration of \(\text{I}_2\) is approximately \(0.128 \, \text{M}\).