The partial pressures of `N_(2)O_(4) "and" NO_(2) "at" 40^(@)C` for the following equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)` are `0.1 "atm and" 0.3` atm respectively. Find `K_(P)` for the reaction.

Explanation

To find the equilibrium constant \( K_p \) for the reaction

\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

given the partial pressures of \( N_2O_4 \) and \( NO_2 \) at 40°C, we will follow these steps:

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

The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the products and reactants. The general formula for \( K_p \) is:

\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

where \( P_{NO_2} \) is the partial pressure of \( NO_2 \) and \( P_{N_2O_4} \) is the partial pressure of \( N_2O_4 \).

### Step 2: Substitute the given values

From the problem, we know:
- \( P_{N_2O_4} = 0.1 \, \text{atm} \)
- \( P_{NO_2} = 0.3 \, \text{atm} \)

Now, we can substitute these values into the \( K_p \) expression:

\[ K_p = \frac{(0.3)^2}{0.1} \]

### Step 3: Calculate \( K_p \)

Now, we will perform the calculations:

1. Calculate \( (0.3)^2 \):
\[ (0.3)^2 = 0.09 \]

2. Substitute this value back into the \( K_p \) expression:
\[ K_p = \frac{0.09}{0.1} \]

3. Perform the division:
\[ K_p = 0.09 \div 0.1 = 0.9 \]

### Final Answer

Thus, the equilibrium constant \( K_p \) for the reaction at 40°C is:

\[ K_p = 0.9 \]

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