The equilibrium constant for the following reaction is `1.6xx10^(5)` at `1024 K`
`H_(2)(g)+Br_(2)(g) hArr 2HBr(g)`
find the equilibrium pressure of all gases if `10.0` bar of `HBr` is introduced into a sealed container at `1024 K`.

Explanation

To solve the problem, we need to find the equilibrium pressures of all gases involved in the reaction:

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

Given:
- Equilibrium constant \( K_p = 1.6 \times 10^5 \) at \( 1024 \, K \)
- Initial pressure of \( HBr = 10.0 \, bar \)

### Step 1: Set up the initial conditions
At the start (t = 0), we have:
- \( P_{HBr} = 10.0 \, bar \)
- \( P_{H2} = 0 \, bar \)
- \( P_{Br2} = 0 \, bar \)

### Step 2: Define the change in pressures
Let \( P \) be the change in pressure of \( HBr \) that reacts to reach equilibrium. Since the stoichiometry of the reaction shows that 2 moles of \( HBr \) are produced for every mole of \( H_2 \) and \( Br_2 \) that reacts, we can express the changes in pressure as follows:
- Change in \( HBr \): \( -2P \)
- Change in \( H2 \): \( +P \)
- Change in \( Br2 \): \( +P \)

### Step 3: Write the equilibrium pressures
At equilibrium, the pressures will be:
- \( P_{HBr} = 10.0 - 2P \)
- \( P_{H2} = P \)
- \( P_{Br2} = P \)

### Step 4: Substitute into the equilibrium expression
The equilibrium constant expression for the reaction in terms of partial pressures is:
\[ K_p = \frac{(P_{HBr})^2}{(P_{H2})(P_{Br2})} \]
Substituting the equilibrium pressures:
\[ 1.6 \times 10^5 = \frac{(10.0 - 2P)^2}{(P)(P)} \]
This simplifies to:
\[ 1.6 \times 10^5 = \frac{(10.0 - 2P)^2}{P^2} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 1.6 \times 10^5 P^2 = (10.0 - 2P)^2 \]

### Step 6: Expand and simplify
Expanding the right side:
\[ 1.6 \times 10^5 P^2 = 100 - 40P + 4P^2 \]
Rearranging gives:
\[ (1.6 \times 10^5 - 4)P^2 + 40P - 100 = 0 \]
This simplifies to:
\[ 1.59996 \times 10^5 P^2 + 40P - 100 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 1.59996 \times 10^5 \)
- \( b = 40 \)
- \( c = -100 \)

Calculating the discriminant:
\[ D = b^2 - 4ac = 40^2 - 4 \times (1.59996 \times 10^5)(-100) \]
Calculating \( P \):
\[ P = \frac{-40 \pm \sqrt{D}}{2 \times 1.59996 \times 10^5} \]

Calculating the positive root gives us \( P \approx 0.0498 \, bar \).

### Step 8: Calculate equilibrium pressures
Now substituting \( P \) back to find the equilibrium pressures:
- \( P_{HBr} = 10.0 - 2(0.0498) \approx 10.0 \, bar \) (remains almost unchanged)
- \( P_{H2} = P \approx 0.0498 \, bar \)
- \( P_{Br2} = P \approx 0.0498 \, bar \)

### Final Answer
- \( P_{HBr} \approx 10.0 \, bar \)
- \( P_{H2} \approx 0.0498 \, bar \)
- \( P_{Br2} \approx 0.0498 \, bar \)