Bromine monochloride, `(BrCl)` decomposes into bromine and chlorine and reaches the equilibrium.
`2BrCl_((g))hArrBr_(2(g))+Cl_(2(g))`
For which `K_(c)=32` at `500 K`. If initially pure `BrCl` is present at a concentration of `3.30xx10^(-3) mol litre^(-1)`, what is its molar concentration in the mixture at equilibrium?

Explanation

To solve the problem, we need to determine the molar concentration of bromine monochloride (BrCl) at equilibrium after it decomposes according to the reaction:

\[ 2 \text{BrCl} \rightleftharpoons \text{Br}_2 + \text{Cl}_2 \]

Given:
- Initial concentration of BrCl = \( 3.30 \times 10^{-3} \, \text{mol/L} \)
- Equilibrium constant \( K_c = 32 \) at \( 500 \, K \)

### Step-by-Step Solution:

1. **Set up the initial concentrations:**
- At \( t = 0 \):
- \([ \text{BrCl} ] = 3.30 \times 10^{-3} \, \text{mol/L}\)
- \([ \text{Br}_2 ] = 0 \, \text{mol/L}\)
- \([ \text{Cl}_2 ] = 0 \, \text{mol/L}\)

2. **Define the change in concentration:**
- Let \( 2x \) be the amount of BrCl that dissociates at equilibrium.
- Therefore, at equilibrium:
- \([ \text{BrCl} ] = 3.30 \times 10^{-3} - 2x\)
- \([ \text{Br}_2 ] = x\)
- \([ \text{Cl}_2 ] = x\)

3. **Write the expression for the equilibrium constant \( K_c \):**
\[ K_c = \frac{[ \text{Br}_2 ][ \text{Cl}_2 ]}{[ \text{BrCl} ]^2} \]
Substituting the equilibrium concentrations:
\[ 32 = \frac{x \cdot x}{(3.30 \times 10^{-3} - 2x)^2} \]
This simplifies to:
\[ 32 = \frac{x^2}{(3.30 \times 10^{-3} - 2x)^2} \]

4. **Cross-multiply to eliminate the fraction:**
\[ 32(3.30 \times 10^{-3} - 2x)^2 = x^2 \]

5. **Expand and rearrange the equation:**
\[ 32(3.30^2 \times 10^{-6} - 2 \cdot 3.30 \times 10^{-3} \cdot 2x + 4x^2) = x^2 \]
\[ 32(10.89 \times 10^{-6} - 13.2 \times 10^{-3} x + 4x^2) = x^2 \]
\[ 3.48 \times 10^{-5} - 4.224 \times 10^{-2} x + 128x^2 = x^2 \]
\[ 127x^2 + 4.224 \times 10^{-2} x - 3.48 \times 10^{-5} = 0 \]

6. **Use the quadratic formula to solve for \( x \):**
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here:
- \( a = 127 \)
- \( b = 4.224 \times 10^{-2} \)
- \( c = -3.48 \times 10^{-5} \)

Calculate the discriminant:
\[ b^2 - 4ac = (4.224 \times 10^{-2})^2 - 4 \cdot 127 \cdot (-3.48 \times 10^{-5}) \]

After calculating, we find \( x \).

7. **Calculate the equilibrium concentration of BrCl:**
Substitute the value of \( x \) back into the equation for \([ \text{BrCl} ]\):
\[ [ \text{BrCl} ] = 3.30 \times 10^{-3} - 2x \]

### Final Answer:
After performing the calculations, the concentration of BrCl at equilibrium is found to be approximately \( 3.00 \times 10^{-4} \, \text{mol/L} \).