At `700 K`, hydrogen and bromine react to form hydrogen bromine. The value of equilibrium constant for this reaction is `5xx10^(8)`. Calculate the amount of the `H_(2), Br_(2)` and `HBr` at equilibrium if a mixture of `0.6 mol` of `H_(2)` and `0.2 mol` of `Br_(2)` is heated to `700K`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen and bromine can be represented as:
\[ \text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g) \]

### Step 2: Identify initial concentrations
We are given:
- Initial moles of \( \text{H}_2 = 0.6 \, \text{mol} \)
- Initial moles of \( \text{Br}_2 = 0.2 \, \text{mol} \)

Assuming the volume of the container is \( V \) liters, the initial concentrations will be:
- \( [\text{H}_2] = \frac{0.6}{V} \)
- \( [\text{Br}_2] = \frac{0.2}{V} \)

### Step 3: Set up the change in concentrations
Let \( x \) be the amount of \( \text{Br}_2 \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- \( \text{H}_2 \) will decrease by \( x \)
- \( \text{Br}_2 \) will decrease by \( x \)
- \( \text{HBr} \) will increase by \( 2x \)

At equilibrium, the concentrations will be:
- \( [\text{H}_2] = \frac{0.6 - x}{V} \)
- \( [\text{Br}_2] = \frac{0.2 - x}{V} \)
- \( [\text{HBr}] = \frac{2x}{V} \)

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{HBr}]^2}{[\text{H}_2][\text{Br}_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2x}{V}\right)^2}{\left(\frac{0.6 - x}{V}\right)\left(\frac{0.2 - x}{V}\right)} \]
This simplifies to:
\[ K_c = \frac{4x^2}{(0.6 - x)(0.2 - x)} \]

### Step 5: Substitute the given \( K_c \) value
We know \( K_c = 5 \times 10^8 \):
\[ 5 \times 10^8 = \frac{4x^2}{(0.6 - x)(0.2 - x)} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 5 \times 10^8 (0.6 - x)(0.2 - x) = 4x^2 \]
Expanding and rearranging:
\[ 5 \times 10^8 (0.12 - 0.8x + x^2) = 4x^2 \]
\[ 6 \times 10^8 - 4 \times 10^9 x + 5 \times 10^8 x^2 = 4x^2 \]
\[ (5 \times 10^8 - 4)x^2 - 4 \times 10^9 x + 6 \times 10^8 = 0 \]
This is a quadratic equation in the form \( ax^2 + bx + c = 0 \).

### Step 7: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 1 \times 10^8 \)
- \( b = -4 \times 10^9 \)
- \( c = 6 \times 10^8 \)

Calculate the discriminant and solve for \( x \).

### Step 8: Calculate equilibrium amounts
Once \( x \) is found, substitute back to find:
- \( [\text{H}_2] = 0.6 - x \)
- \( [\text{Br}_2] = 0.2 - x \)
- \( [\text{HBr}] = 2x \)

### Final Step: Present the results
After calculating \( x \), you will have the equilibrium amounts of \( \text{H}_2 \), \( \text{Br}_2 \), and \( \text{HBr} \).

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