In the reaction `PCl_5(g)⇔PCl_3(g)+Cl_2(g) `the equilibrium concentration of PCl5 and PCl3 are 0.4 and 0.2 mole/litre respectively. If the value of KC is 0.5 what is concentration of `Cl_2` in moles/litre ?

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the equilibrium expression
For the reaction:
\[ PCl_5(g) ⇔ PCl_3(g) + Cl_2(g) \]

The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 2: Substitute the known values
From the problem, we know:
- The concentration of \( PCl_5 \) is 0.4 mol/L
- The concentration of \( PCl_3 \) is 0.2 mol/L
- The value of \( K_c \) is 0.5

Substituting these values into the equilibrium expression gives:
\[ 0.5 = \frac{(0.2)[Cl_2]}{0.4} \]

### Step 3: Solve for the concentration of \( Cl_2 \)
To find the concentration of \( Cl_2 \), we can rearrange the equation:
\[ 0.5 = \frac{0.2 \cdot [Cl_2]}{0.4} \]

Multiplying both sides by 0.4:
\[ 0.5 \cdot 0.4 = 0.2 \cdot [Cl_2] \]
\[ 0.2 = 0.2 \cdot [Cl_2] \]

Now, divide both sides by 0.2:
\[ [Cl_2] = \frac{0.2}{0.2} \]
\[ [Cl_2] = 1 \text{ mol/L} \]

### Conclusion
The concentration of \( Cl_2 \) is 1 mol/L.

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