A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\) . After equilibrium is reached the total pressure is 1.5 atm and 16\(\%\) (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( has dissociated to \)\mathrm{NO}_{2}(g) .$ a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C} .\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g) .\) c. What percentage (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)$ is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{atm} ) ?\)

Step by step solution

01

Write the Reaction Equilibrium Expression

The chemical reaction can be written as follows: \( N_2O_4(g) \leftrightarrow 2NO_2(g) \) The equilibrium constant expression (\(K_p\)) for this reaction can be written as: \( K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \), where \(P_{NO_2}\) is the partial pressure of NO2 and \(P_{N_2O_4}\) is the partial pressure of N2O4.

02

Calculate the Partial Pressures under Initial Conditions

Given 16% of N2O4 dissociates, so in every 100 moles of N2O4 initially present, 16 moles dissociate into 32 moles of NO2. Hence when equilibrium is established: For \(N2O4: P_{N2O_4} = 1.5 \times \frac{100 - 16}{100 + 32}\) atm = 0.765 atm For \(NO2: P_{NO_2} = 1.5 \times \frac{32}{100 + 32}\) atm = 0.735 atm

03

Calculate the Value of \(K_p\) under Initial Conditions

Substitute \(P_{N2O_4}\) and \(P_{NO_2}\) values into the \(K_p\) equation : \( K_p = \frac{(0.735)^2}{0.765} = 0.70 \)

04

Calculate the Partial Pressures after Volume Increase

The total pressure is given to have dropped to 1 atm due to the increase in volume. According to Le Chatelier's principle, the reaction will shift in the direction in which more gaseous moles are present. Thus, volume increase will cause more dissociation of N2O4. Let the extra percentage of N2O4 dissociated be x. So, for \(N2O4: P_{N2O_4} = 1 \times \frac{100 - (16 + x)}{100 + 32 + 2x}\) For \(NO2: P_{NO_2} = 1 \times \frac{32 + 2x}{100 + 32 + 2x}\)

05

Evaluate Percentage of Additional N2O4 Dissociated

Apply \(K_p\) value to the new equilibrium equation: \( 0.70 = \frac{(\frac{32 + 2x}{132 + 2x})^2}{\frac{84 - x}{132 + 2x}}\) Solving this yields x = 27. Therefore, the new pressures when equilibrium is reestablished: \( P_{N2O4} = 0.35 \) atm, and \( P_{NO2} = 0.65 \) atm

06

Calculate the Percentage of Original N2O4 Dissociated

Percentage dissociated = (Initial moles - Remaining moles) / Initial moles * 100 = [(100 - (100 - (16 + 27))] / 100 * 100 = 43\% of the original N2O4 has dissociated at the new equilibrium position.

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