In a 1-litre vessel at 1000 K are introduced 0.1 mole each of NO and `Br_2` and 0.01 mole of NOBr
`2NO (g)+Br_(2) (g) Leftrightarrow 2NOBr (g), K_(c)=1.32 xx 10^(-2)`
Determine the direction of the net reaction and calculate the partial pressure of NOBr in the vessel at equilibrium.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2NO(g) + Br_2(g) \leftrightarrow 2NOBr(g) \]

### Step 2: Identify initial concentrations
Given:
- Initial moles of \( NO = 0.1 \, \text{mol} \)
- Initial moles of \( Br_2 = 0.1 \, \text{mol} \)
- Initial moles of \( NOBr = 0.01 \, \text{mol} \)

Since the volume of the vessel is 1 liter, the initial concentrations are:
- \([NO] = 0.1 \, \text{mol/L}\)
- \([Br_2] = 0.1 \, \text{mol/L}\)
- \([NOBr] = 0.01 \, \text{mol/L}\)

### Step 3: Calculate the reaction quotient \( Q_c \)
The expression for the equilibrium constant \( K_c \) is:
\[ K_c = \frac{[NOBr]^2}{[NO]^2 \cdot [Br_2]} \]

Substituting the initial concentrations into the expression:
\[ Q_c = \frac{(0.01)^2}{(0.1)^2 \cdot (0.1)} = \frac{0.0001}{0.01} = 0.01 \]

### Step 4: Compare \( Q_c \) with \( K_c \)
Given \( K_c = 1.32 \times 10^{-2} \):
- Since \( Q_c (0.01) < K_c (1.32 \times 10^{-2}) \), the reaction will shift to the right (towards the products) to reach equilibrium.

### Step 5: Set up the equilibrium expression
Let \( x \) be the change in moles of \( NOBr \) at equilibrium. The changes in concentrations will be:
- \([NO] = 0.1 - x\)
- \([Br_2] = 0.1 - \frac{x}{2}\)
- \([NOBr] = 0.01 + x\)

### Step 6: Write the equilibrium expression in terms of \( x \)
At equilibrium:
\[ K_c = \frac{(0.01 + x)^2}{(0.1 - x)^2 \cdot (0.1 - \frac{x}{2})} \]

Substituting \( K_c = 1.32 \times 10^{-2} \):
\[ 1.32 \times 10^{-2} = \frac{(0.01 + x)^2}{(0.1 - x)^2 \cdot (0.1 - \frac{x}{2})} \]

### Step 7: Make assumptions to simplify calculations
Assuming \( x \) is small compared to 0.1, we can simplify:
- \( 0.1 - x \approx 0.1 \)
- \( 0.1 - \frac{x}{2} \approx 0.1 \)

Thus, the equation simplifies to:
\[ 1.32 \times 10^{-2} = \frac{(0.01 + x)^2}{(0.1)^2 \cdot (0.1)} \]

### Step 8: Solve for \( x \)
Rearranging gives:
\[ 1.32 \times 10^{-2} = \frac{(0.01 + x)^2}{0.01} \]
\[ 1.32 \times 10^{-4} = (0.01 + x)^2 \]

Taking the square root:
\[ 0.0115 \approx 0.01 + x \]
Thus:
\[ x \approx 0.0015 \]

### Step 9: Calculate equilibrium concentrations
Now, substituting \( x \) back:
- \([NOBr] = 0.01 + 0.0015 = 0.0115 \, \text{mol/L}\)

### Step 10: Calculate partial pressure of \( NOBr \) at equilibrium
Using the ideal gas law:
\[ P = \frac{nRT}{V} \]
Where:
- \( n = 0.0115 \, \text{mol} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 1000 \, K \)
- \( V = 1 \, L \)

Calculating:
\[ P = \frac{0.0115 \times 0.0821 \times 1000}{1} \approx 0.946 \, \text{atm} \]

### Final Result
The partial pressure of \( NOBr \) at equilibrium is approximately \( 0.946 \, \text{atm} \).

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