At equilibrium for the reaction:
`H_2(g)+Br_2(g) hArr 2HBr(g)`
a 10 L reaction vessel was found to contain `2.5 xx 10^(-3)` mol of `H_2`, 0.150 mol of HBr and `2.8 xx 10^(-3)` mol of `Br_2`. What is the value of K at this temperature ?

Explanation

To find the equilibrium constant \( K \) for the reaction

\[ H_2(g) + Br_2(g) \rightleftharpoons 2 HBr(g) \]

given the concentrations of the reactants and products at equilibrium, we follow these steps:

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

The equilibrium constant \( K \) for the reaction is given by the formula:

\[ K = \frac{[HBr]^2}{[H_2][Br_2]} \]

where \([HBr]\), \([H_2]\), and \([Br_2]\) are the molar concentrations of HBr, H2, and Br2 at equilibrium.

### Step 2: Calculate the concentrations of the reactants and products

We need to convert the number of moles of each substance into concentrations (molarity), using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Given:
- Volume of the reaction vessel = 10 L
- Moles of \( H_2 = 2.5 \times 10^{-3} \) mol
- Moles of \( HBr = 0.150 \) mol
- Moles of \( Br_2 = 2.8 \times 10^{-3} \) mol

Now, we calculate the concentrations:

1. **Concentration of \( H_2 \)**:
\[ [H_2] = \frac{2.5 \times 10^{-3}}{10} = 2.5 \times 10^{-4} \, \text{mol/L} \]

2. **Concentration of \( HBr \)**:
\[ [HBr] = \frac{0.150}{10} = 1.5 \times 10^{-2} \, \text{mol/L} \]

3. **Concentration of \( Br_2 \)**:
\[ [Br_2] = \frac{2.8 \times 10^{-3}}{10} = 2.8 \times 10^{-4} \, \text{mol/L} \]

### Step 3: Substitute the concentrations into the equilibrium constant expression

Now we substitute the concentrations we calculated into the \( K \) expression:

\[ K = \frac{(1.5 \times 10^{-2})^2}{(2.5 \times 10^{-4})(2.8 \times 10^{-4})} \]

### Step 4: Calculate the value of \( K \)

1. Calculate the numerator:
\[ (1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4} \]

2. Calculate the denominator:
\[ (2.5 \times 10^{-4})(2.8 \times 10^{-4}) = 7.0 \times 10^{-8} \]

3. Now, substitute these values into the \( K \) expression:
\[ K = \frac{2.25 \times 10^{-4}}{7.0 \times 10^{-8}} = 3.21 \times 10^{3} \]

### Final Result

Thus, the value of the equilibrium constant \( K \) at this temperature is:

\[ K = 3.21 \times 10^{3} \]

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