The equilibrium composition for the reaction is
`{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.20,,0.10,,0.40 mol L^(-1)):}`
What will be the equilibrium concentration of `PCl_(5)` on adding `0.10 mol` of `Cl_(2)` at the same temperature?

Explanation

To solve the problem, we need to determine the equilibrium concentration of \( PCl_5 \) after adding \( 0.10 \, \text{mol} \) of \( Cl_2 \) to the system. Let's break it down step by step.

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ PCl_3 + Cl_2 \rightleftharpoons PCl_5 \]

### Step 2: Identify Initial Concentrations
From the problem, we have the initial equilibrium concentrations:
- \( [PCl_3] = 0.20 \, \text{mol L}^{-1} \)
- \( [Cl_2] = 0.10 \, \text{mol L}^{-1} \)
- \( [PCl_5] = 0.40 \, \text{mol L}^{-1} \)

### Step 3: Calculate the Reaction Quotient \( Q_c \)
The reaction quotient \( Q_c \) is given by the formula:
\[ Q_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} \]
Substituting the initial concentrations:
\[ Q_c = \frac{0.40}{0.20 \times 0.10} = \frac{0.40}{0.02} = 20 \]

### Step 4: Add \( 0.10 \, \text{mol} \) of \( Cl_2 \)
After adding \( 0.10 \, \text{mol} \) of \( Cl_2 \), the new concentration of \( Cl_2 \) becomes:
\[ [Cl_2] = 0.10 + 0.10 = 0.20 \, \text{mol L}^{-1} \]
Now, the concentrations are:
- \( [PCl_3] = 0.20 \, \text{mol L}^{-1} \)
- \( [Cl_2] = 0.20 \, \text{mol L}^{-1} \)
- \( [PCl_5] = 0.40 \, \text{mol L}^{-1} \)

### Step 5: Calculate the New Reaction Quotient \( Q_c \)
Now, we calculate the new \( Q_c \):
\[ Q_c = \frac{0.40}{0.20 \times 0.20} = \frac{0.40}{0.04} = 10 \]

### Step 6: Compare \( Q_c \) and \( K_c \)
Since the original \( K_c \) was \( 20 \) and the new \( Q_c \) is \( 10 \), we have:
\[ Q_c < K_c \]
This indicates that the reaction will shift to the right to reach a new equilibrium.

### Step 7: Set Up the Equilibrium Expression
Let \( x \) be the amount of \( PCl_3 \) and \( Cl_2 \) that react to form \( PCl_5 \) at the new equilibrium. The new concentrations will be:
- \( [PCl_3] = 0.20 - x \)
- \( [Cl_2] = 0.20 - x \)
- \( [PCl_5] = 0.40 + x \)

### Step 8: Write the Equilibrium Expression
Using the equilibrium expression:
\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = 20 \]
Substituting the expressions for the concentrations:
\[ 20 = \frac{0.40 + x}{(0.20 - x)(0.20 - x)} \]

### Step 9: Solve for \( x \)
Expanding and rearranging the equation:
\[ 20(0.20 - x)^2 = 0.40 + x \]
\[ 20(0.04 - 0.4x + x^2) = 0.40 + x \]
\[ 0.8 - 8x + 20x^2 = 0.40 + x \]
\[ 20x^2 - 9x + 0.4 = 0 \]

### Step 10: Use the Quadratic Formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 20 \times 0.4}}{2 \times 20} \]
Calculating the discriminant:
\[ x = \frac{9 \pm \sqrt{81 - 32}}{40} \]
\[ x = \frac{9 \pm \sqrt{49}}{40} \]
\[ x = \frac{9 \pm 7}{40} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{16}{40} = 0.4 \)
2. \( x = \frac{2}{40} = 0.05 \)

Since \( x \) must be less than \( 0.20 \) (the initial concentration of \( Cl_2 \)), we take \( x = 0.05 \).

### Step 11: Calculate the Final Concentration of \( PCl_5 \)
Now, substituting \( x \) back into the expression for \( [PCl_5] \):
\[ [PCl_5] = 0.40 + 0.05 = 0.45 \, \text{mol L}^{-1} \]

### Final Answer
The equilibrium concentration of \( PCl_5 \) after adding \( 0.10 \, \text{mol} \) of \( Cl_2 \) is:
\[ [PCl_5] = 0.45 \, \text{mol L}^{-1} \]