Question

At 298 K, the value of K\(_p\) for N\(_2\)O\(_4\)(g) \(\rightleftharpoons\) 2NO\(_2\)(g) is 0.113 atm. The partial pressure of N\(_2\)O\(_4\) at equilibrium is 0.2 atm. What is the partial pressure (in atm) of NO\(_2\) at equilibrium ?

• A. \(0.05 \)
• B. \(0.075 \)
• C. \(0.30 \)
• D. \(0.15 \)

Hint:

To solve problems involving \(K_p\): 1. Write the balanced chemical equation. 2. Formulate the \(K_p\) expression: Products raised to their stoichiometric coefficients divided by reactants raised to theirs. Remember that partial pressures are used for gases. 3. Substitute known values: Plug in the given \(K_p\) value and equilibrium partial pressures. 4. Solve for the unknown: Algebraically rearrange and solve for the desired partial pressure.

Answer 1

The Correct Option is D

Step 1: Write down the equilibrium expression for K\(_p\).
The given reversible reaction in the gaseous phase is:
\[ \operatorname{N}_2\operatorname{O}_4(\operatorname{g}) \rightleftharpoons 2\operatorname{NO}_2(\operatorname{g}) \] For a general reversible reaction \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\), the equilibrium constant in terms of partial pressures, K\(_p\), is given by: \[ K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \] For the given reaction, the expression for K\(_p\) is: \[ K_p = \frac{(P_{\operatorname{NO}_2})^2}{P_{\operatorname{N}_2\operatorname{O}_4}} \] where \(P_{\operatorname{NO}_2}\) is the partial pressure of NO\(_2\) at equilibrium and \(P_{\operatorname{N}_2\operatorname{O}_4}\) is the partial pressure of N\(_2\)O\(_4\) at equilibrium. 
Step 2: Identify the given values from the problem statement.
We are given:
Equilibrium constant, \(K_p = 0.113 \operatorname{atm}\)
Partial pressure of N\(_2\)O\(_4\) at equilibrium, \(P_{\operatorname{N}_2\operatorname{O}_4} = 0.2 \operatorname{atm}\) We need to find the partial pressure of NO\(_2\) at equilibrium, \(P_{\operatorname{NO}_2}\). 
Step 3: Substitute the known values into the K\(_p\) expression and solve for the unknown partial pressure.
Substitute the given values into the \(K_p\) expression: \[ 0.113 = \frac{(P_{\operatorname{NO}_2})^2}{0.2} \] Now, rearrange the equation to solve for \((P_{\operatorname{NO}_2})^2\): \[ (P_{\operatorname{NO}_2})^2 = 0.113 \times 0.2 \] \[ (P_{\operatorname{NO}_2})^2 = 0.0226 \] To find \(P_{\operatorname{NO}_2}\), take the square root of both sides: \[ P_{\operatorname{NO}_2} = \sqrt{0.0226} \] \[ P_{\operatorname{NO}_2} \approx 0.15033 \] 
Step 4: Round the answer and compare with the given options.
Rounding the calculated value to two or three significant figures, we get: \(P_{\operatorname{NO}_2} \approx 0.15 \operatorname{atm}\) This value matches option (4). 
The final answer is $\boxed{0.15}$.