At equilibrium of the reaction `PCl_5(g)⇔PCl_3(g)+Cl_2(g) `the concentrations of `PCl_5(g) `and `PCl_3(g)` are 0.2 and 0.1 moles/lit.respectively `K_c` is 0.05. The equilibrium concentration of `Cl_2`

Answer

To find the equilibrium concentration of \( Cl_2 \) in the reaction

\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction can be expressed as:

\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 2: Substitute the known values into the equation

From the problem, we know:

- \( [PCl_5] = 0.2 \, \text{mol/L} \)
- \( [PCl_3] = 0.1 \, \text{mol/L} \)
- \( K_c = 0.05 \)

Substituting these values into the \( K_c \) expression gives:

\[ 0.05 = \frac{(0.1)[Cl_2]}{0.2} \]

### Step 3: Solve for the concentration of \( Cl_2 \)

Now we can rearrange the equation to solve for \( [Cl_2] \):

\[ 0.05 = \frac{0.1 \cdot [Cl_2]}{0.2} \]

Multiplying both sides by \( 0.2 \):

\[ 0.05 \cdot 0.2 = 0.1 \cdot [Cl_2] \]

Calculating the left side:

\[ 0.01 = 0.1 \cdot [Cl_2] \]

Now, divide both sides by \( 0.1 \):

\[ [Cl_2] = \frac{0.01}{0.1} = 0.1 \, \text{mol/L} \]

### Conclusion

The equilibrium concentration of \( Cl_2 \) is \( 0.1 \, \text{mol/L} \).

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