NO and `Br_(2)` at initial pressures of `98.4` and `41.3` torr respectively were allowed react at `300 K`. At equilibrium the total pressure was `110.5` torr. Calculate the value of equilibrium constant, `K_(p)` and the standard free energy change at `300 K` for the reaction:
`2NO(g)+Br_(2)(g) hArr 2NOBr(g)`

Explanation

To solve the problem step by step, we will follow the reaction given and calculate the equilibrium constant \( K_p \) and the standard free energy change \( \Delta G \) at 300 K.

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ 2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g) \]

### Step 2: Define initial pressures
The initial pressures are:
- \( P_{\text{NO}} = 98.4 \, \text{torr} \)
- \( P_{\text{Br}_2} = 41.3 \, \text{torr} \)

At the start, the pressure of the product \( \text{NOBr} \) is 0 torr.

### Step 3: Set up the change in pressures
Let \( x \) be the change in pressure of \( \text{Br}_2 \) that reacts. Therefore:
- The change in pressure for \( \text{NO} \) will be \( 2x \) (since 2 moles of NO react).
- The change in pressure for \( \text{NOBr} \) will be \( 2x \) (since 2 moles of NOBr are produced).

At equilibrium:
- Pressure of \( \text{NO} = 98.4 - 2x \)
- Pressure of \( \text{Br}_2 = 41.3 - x \)
- Pressure of \( \text{NOBr} = 2x \)

### Step 4: Write the total pressure equation
The total pressure at equilibrium is given as:
\[ P_{\text{total}} = (98.4 - 2x) + (41.3 - x) + (2x) = 110.5 \, \text{torr} \]

### Step 5: Simplify the equation
Combine the terms:
\[ 98.4 - 2x + 41.3 - x + 2x = 110.5 \]
This simplifies to:
\[ 139.7 - x = 110.5 \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x = 139.7 - 110.5 = 29.2 \, \text{torr} \]

### Step 7: Calculate equilibrium pressures
Now substitute \( x \) back to find the equilibrium pressures:
- \( P_{\text{NO}} = 98.4 - 2(29.2) = 40 \, \text{torr} \)
- \( P_{\text{Br}_2} = 41.3 - 29.2 = 12.1 \, \text{torr} \)
- \( P_{\text{NOBr}} = 2(29.2) = 58.4 \, \text{torr} \)

### Step 8: Calculate the equilibrium constant \( K_p \)
The expression for \( K_p \) is:
\[ K_p = \frac{(P_{\text{NOBr}})^2}{(P_{\text{NO}})^2 \cdot (P_{\text{Br}_2})} \]

Substituting the equilibrium pressures:
\[ K_p = \frac{(58.4)^2}{(40)^2 \cdot (12.1)} \]

Calculating:
\[ K_p = \frac{3414.56}{1600 \cdot 12.1} = \frac{3414.56}{19360} \approx 0.176 \]

### Step 9: Convert \( K_p \) to atm
To convert torr to atm, we use the conversion factor \( 1 \, \text{atm} = 760 \, \text{torr} \):
\[ K_p = 0.176 \times \frac{760}{1} \approx 133.76 \, \text{atm} \]

### Step 10: Calculate the standard free energy change \( \Delta G \)
Using the formula:
\[ \Delta G = -2.303 \cdot R \cdot T \cdot \log K_p \]

Where:
- \( R = 8.314 \, \text{J/mol K} \)
- \( T = 300 \, \text{K} \)

Substituting the values:
\[ \Delta G = -2.303 \cdot 8.314 \cdot 300 \cdot \log(133.76) \]

Calculating:
\[ \Delta G \approx -2.303 \cdot 8.314 \cdot 300 \cdot 2.127 \]
\[ \Delta G \approx -12.22 \, \text{kJ/mol} \]

### Final Answers
- \( K_p \approx 133.76 \, \text{atm} \)
- \( \Delta G \approx -12.22 \, \text{kJ/mol} \)