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Chapter 13: Problem 108

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 \(1.0 \mathrm{~atm}\) (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 \(=1.00 \mathrm{~atm}\) )?

Short Answer

Expert verified
The value of \(K_p\) for the dissociation of N2O4 at 25°C is 0.152. When the total pressure is reduced to 1.0 atm, the equilibrium pressures of N2O4(g) and NO2(g) are 0.8831 atm and 0.6182 atm, respectively. At this new equilibrium position, 18.6% of the original N2O4(g) is dissociated.

Step by step solution

01

Set up the ICE table

Initially, we are given that 16% of the original N2O4(g) has dissociated at equilibrium. Let the initial moles of N2O4 be 'x' moles. Then at equilibrium, 0.16x moles have dissociated. Since 2 moles of NO2 are formed for every mole of N2O4 dissociated, at equilibrium, there are 2 * (0.16x) = 0.32x moles of NO2. For pressure, since the mole fraction is directly proportional to partial pressure, Initial pressure = P Equilibrium pressure of N2O4 = 0.84 * P (since initially, the mole fraction is 1) Equilibrium pressure of NO2 = 0.32 * P Total pressure at equilibrium = 1.5 atm So, P * (0.84 + 0.32) = 1.5
02

Solve for P and Kp

Solve for the pressure P: \(P(0.84 + 0.32) = 1.5\) \(P = \frac{1.5}{(1.16)}\) \(P = 1.2931 \thinspace atm\) Now, we can find the value of Kp using the expression: \(K_{p} = \frac{(0.32 \thinspace P)^2}{(0.84 \thinspace P)}\) \(K_{p} = \frac{(0.32)^2 \times (1.2931)^2}{(0.84) \times (1.2931)}\) \(K_{p} = 0.152\) So the Kp value for the dissociation of N2O4 is 0.152. **b. Calculate the equilibrium pressure of N2O4(g) and NO2(g) when the total pressure is reduced to 1.0 atm.**
03

Set up the new ICE table

We now have to find the new equilibrium pressures of N2O4 and NO2 when the total pressure is changed to 1.0 atm. We can use the initial equilibrium pressures of N2O4 and NO2, which are 0.84 * P and 0.32 * P, respectively. Change in pressure of N2O4 = -x Change in pressure of NO2 = +2x New equilibrium pressure of N2O4 = (0.84 * P) - x New equilibrium pressure of NO2 = (0.32 * P) + 2x
04

Solve for new pressure changes

Use the new Kp value and the reaction quotient, Q, to solve for x: Q = \(\frac{((0.32P) + 2x)^2}{((0.84P) - x)}\) Kp = Q (at equilibrium) 0.152 = \(\frac{((0.32 \times 1.2931) + 2x)^2}{((0.84 \times 1.2931) - x)}\) Solve for x to get x = 0.1991 atm
05

Calculate the new equilibrium pressures

Use the value of x to find the new equilibrium pressures: New equilibrium pressure of N2O4 = (0.84 * P) - x = (0.84 * 1.2931) - 0.1991 = 0.8831 atm New equilibrium pressure of NO2 = (0.32 * P) + 2x = (0.32 * 1.2931) + 2 * 0.1991 = 0.6182 atm **c. What percentage (by moles) of the original N2O4(g) is dissociated at the new equilibrium position (total pressure = 1.00 atm)?**
06

Calculate the percentage dissociation

We can find the percentage dissociation by comparing the initial and new equilibrium pressures: Initial equilibrium pressure of N2O4 = 0.84 * P = 0.84 * 1.2931 = 1.085 New equilibrium pressure of N2O4 = 0.8831 atm Percentage dissociation = \(\frac{Initial \thinspace equilibrium \thinspace pressure \thinspace of \thinspace N_{2}O_{4} - New \thinspace equilibrium \thinspace pressure \thinspace of \thinspace N_{2}O_{4}}{Initial \thinspace equilibrium \thinspace pressure \thinspace of \thinspace N_{2}O_{4}}\) × 100 Percentage dissociation = \(\frac{1.085 - 0.8831}{1.085}\) × 100 = 18.6% At the new equilibrium position, 18.6% of the original N2O4(g) is dissociated.

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