What is the equilibrium concentration of each of the substance in the equilibrium when the initial concentration of `ICl` was `0.78 M`?
`2ICl(g) hArr I_(2)(g)+Cl_(2)(g), K_(c)=0.14`

Explanation

To find the equilibrium concentrations of each substance in the reaction \(2 \text{ICl}(g) \rightleftharpoons \text{I}_2(g) + \text{Cl}_2(g)\) with an initial concentration of \(\text{ICl} = 0.78 \, M\) and \(K_c = 0.14\), we can follow these steps:

### Step 1: Set up the initial concentrations
At the start of the reaction (t = 0):
- \([\text{ICl}] = 0.78 \, M\)
- \([\text{I}_2] = 0 \, M\)
- \([\text{Cl}_2] = 0 \, M\)

### Step 2: Define the change in concentrations
Let \(x\) be the amount of \(\text{ICl}\) that dissociates at equilibrium. The changes in concentration will be:
- \([\text{ICl}] = 0.78 - 2x\)
- \([\text{I}_2] = x\)
- \([\text{Cl}_2] = x\)

### Step 3: Write the expression for \(K_c\)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2} \]
Substituting the equilibrium concentrations into this expression gives:
\[ 0.14 = \frac{x \cdot x}{(0.78 - 2x)^2} \]
This simplifies to:
\[ 0.14 = \frac{x^2}{(0.78 - 2x)^2} \]

### Step 4: Solve for \(x\)
Cross-multiplying gives:
\[ 0.14(0.78 - 2x)^2 = x^2 \]
Expanding the left side:
\[ 0.14(0.6084 - 3.12x + 4x^2) = x^2 \]
This simplifies to:
\[ 0.085176 - 0.4368x + 0.56x^2 = x^2 \]
Rearranging gives:
\[ -0.44x^2 - 0.4368x + 0.085176 = 0 \]
Multiplying through by -1:
\[ 0.44x^2 + 0.4368x - 0.085176 = 0 \]

### Step 5: Use the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- \(a = 0.44\)
- \(b = 0.4368\)
- \(c = -0.085176\)

Calculating the discriminant:
\[ b^2 - 4ac = (0.4368)^2 - 4(0.44)(-0.085176) \]
Calculating gives:
\[ 0.1915 + 0.1507 = 0.3422 \]
Now substituting back into the quadratic formula:
\[ x = \frac{-0.4368 \pm \sqrt{0.3422}}{2 \times 0.44} \]
Calculating the square root and the values for \(x\):
\[ x \approx 0.166 \]

### Step 6: Calculate equilibrium concentrations
Now we can find the equilibrium concentrations:
- \([\text{ICl}] = 0.78 - 2(0.166) = 0.78 - 0.332 = 0.448 \, M\)
- \([\text{I}_2] = x = 0.166 \, M\)
- \([\text{Cl}_2] = x = 0.166 \, M\)

### Final Answer
The equilibrium concentrations are:
- \([\text{ICl}] = 0.448 \, M\)
- \([\text{I}_2] = 0.166 \, M\)
- \([\text{Cl}_2] = 0.166 \, M\)