Phosphorus pentachloride decomposes according to the reaction :
`PCl_5(g) hArr PCl_3(g) + Cl_2(g)`
1.50 mol of `PCl_5` is placed in a 500 mL of closed reaction vessel and allowed to reach equilibrium with its decomposition products, phosphorus trichloride and chlorine at 523 K. Equilibrium constant `K_c` is 1.80. What is the composition of mixture of equilibrium ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of phosphorus pentachloride can be represented as:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

### Step 2: Determine the initial concentration of PCl5
We are given that 1.50 moles of PCl5 is placed in a 500 mL vessel. To find the concentration, we use the formula:
\[ \text{Concentration (C)} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Converting 500 mL to liters:
\[ 500 \, \text{mL} = 0.500 \, \text{L} \]
Now, we can calculate the concentration:
\[ C = \frac{1.50 \, \text{mol}}{0.500 \, \text{L}} = 3.00 \, \text{mol/L} \]

### Step 3: Set up an ICE table
We will set up an ICE (Initial, Change, Equilibrium) table to track the concentrations of the reactants and products.

| Species | Initial (M) | Change (M) | Equilibrium (M) |
|--------------|-------------|------------|------------------|
| PCl5 | 3.00 | -x | 3.00 - x |
| PCl3 | 0 | +x | x |
| Cl2 | 0 | +x | x |

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]
Substituting the equilibrium concentrations from the ICE table:
\[ K_c = \frac{x \cdot x}{3.00 - x} = \frac{x^2}{3.00 - x} \]

### Step 5: Substitute the value of \( K_c \) and solve for \( x \)
We know that \( K_c = 1.80 \):
\[ 1.80 = \frac{x^2}{3.00 - x} \]
Cross-multiplying gives:
\[ 1.80(3.00 - x) = x^2 \]
Expanding this:
\[ 5.40 - 1.80x = x^2 \]
Rearranging to form a quadratic equation:
\[ x^2 + 1.80x - 5.40 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 1.80, c = -5.40 \):
\[ b^2 - 4ac = (1.80)^2 - 4(1)(-5.40) = 3.24 + 21.60 = 24.84 \]
Now substituting into the formula:
\[ x = \frac{-1.80 \pm \sqrt{24.84}}{2} \]
Calculating \( \sqrt{24.84} \approx 4.98 \):
\[ x = \frac{-1.80 \pm 4.98}{2} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{3.18}{2} \approx 1.59 \)
2. \( x = \frac{-6.78}{2} \) (not physically meaningful since concentration cannot be negative)

Thus, \( x \approx 1.59 \).

### Step 7: Calculate equilibrium concentrations
Now we can find the equilibrium concentrations:
- For PCl5:
\[ [\text{PCl}_5] = 3.00 - x = 3.00 - 1.59 = 1.41 \, \text{mol/L} \]
- For PCl3:
\[ [\text{PCl}_3] = x = 1.59 \, \text{mol/L} \]
- For Cl2:
\[ [\text{Cl}_2] = x = 1.59 \, \text{mol/L} \]

### Final Answer
The composition of the mixture at equilibrium is:
- \([\text{PCl}_5] = 1.41 \, \text{mol/L}\)
- \([\text{PCl}_3] = 1.59 \, \text{mol/L}\)
- \([\text{Cl}_2] = 1.59 \, \text{mol/L}\)