For the reaction `NOBr(g) hArr NO(g +1//2Br_2(g), K_P` = 0.15 atm at `90^@C`. If 0.5 atm of NOBr, 0.40 atm of NO and 0.2 atm of `Br_2` are mixed at this temperature, will `Br_2` be consumed or formed.

Explanation

To determine whether Br₂ will be consumed or formed in the reaction \( \text{NOBr}(g) \rightleftharpoons \text{NO}(g) + \frac{1}{2} \text{Br}_2(g) \) at equilibrium, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_p \).

For the reaction given, the equilibrium constant \( K_p \) is expressed as:

\[
K_p = \frac{P_{\text{NO}} \cdot P_{\text{Br}_2}^{1/2}}{P_{\text{NOBr}}}
...