The partial pressures of `NO`, `Br_2`, and `NOBr` in a flask at `25^@C` are `0.01, 0.1`, and `0.04atm`, respectively. If the equilibrium constant at `25^@C` for the reaction
`2NO(g)+Br_2(g)hArr 2NOBr(g)`
is equal to `160atm^(-1)`, then we can say that

Options

A. there is equilibrium in the flask
B. there reaction will proceed in the forward
C. the reaction will proceed in the backward direction
D. the partial pressure of NOBr finally will be 0.05 atm

Answer

To solve the problem, we need to determine whether the reaction is at equilibrium or if it will proceed in the forward or backward direction. We will do this by calculating the reaction quotient (Q) and comparing it with the equilibrium constant (K).

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ 2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g) \]

2. **Identify the partial pressures given:**
- \( P_{NO} = 0.01 \, \text{atm} \)
- \( P_{Br_2} = 0.1 \, \text{atm} \)
- \( P_{NOBr} = 0.04 \, \text{atm} \)

3. **Write the expression for the reaction quotient (Q):**
The reaction quotient \( Q \) is given by:
\[ Q = \frac{(P_{NOBr})^2}{(P_{NO})^2 \cdot (P_{Br_2})} \]

4. **Substitute the values into the expression for Q:**
\[ Q = \frac{(0.04)^2}{(0.01)^2 \cdot (0.1)} \]

5. **Calculate \( Q \):**
\[ Q = \frac{0.0016}{0.0001 \cdot 0.1} = \frac{0.0016}{0.00001} = 160 \]

6. **Compare Q with K:**
The equilibrium constant \( K \) is given as \( 160 \, \text{atm}^{-1} \).
- Since \( Q = K \), we conclude that the system is at equilibrium.

### Conclusion:
Since the reaction quotient \( Q \) is equal to the equilibrium constant \( K \), we can say that the reaction is at equilibrium in the flask.

### Final Answer:
The correct option is that there is equilibrium in the flask.

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