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 = I.00 atm)?

Step by step solution

01

Write the chemical equation

The dissociation reaction of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) to \(\mathrm{NO}_{2}(g)\) is given by the following equation: \[2\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)\]

02

Write the Kp expression

The expression for \(K_{\mathrm{p}}\) for the given reaction can be written as: \[K_{p}=\frac{[\mathrm{NO}_{2}]^{2}}{[\mathrm{N}_{2}\mathrm{O}_{4}]}\] Where [NO2] and [N2O4] are the equilibrium concentrations of nitric oxide and dinitrogen tetraoxide, respectively.

03

Calculate the partial pressures of the gases at equilibrium

We are given that \(16\%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated into \(\mathrm{NO}_{2}(g)\). So, there will be a decrease of 0.16 in the moles of N2O4. At the same time, there will be an increase of 2 times the decrease in moles of N2O4, so 2*0.16 = 0.32 moles of NO2 form. We are also given the total pressure at equilibrium: 1.5 atm. Considering the mole fractions, we can calculate the partial pressures of each gas as follows: \[P_{N_{2}O_{4}} = 1.5 \times (1-0.16) = 1.5 \times 0.84 = 1.26 \mathrm{~atm}\] \[P_{NO_{2}} = 1.5 \times 2 \times 0.16 = 1.5 \times 0.32 = 0.48 \mathrm{~atm}\]

04

Calculate the value of Kp

Now, we can calculate the value of \(K_{p}\) using the partial pressures derived in the previous step: \[K_{p}=\frac{[\mathrm{NO}_{2}]^{2}}{[\mathrm{N}_{2}\mathrm{O}_{4}]} = \frac{(0.48)^{2}}{1.26} = 0.183\] So, the value of \(K_{p}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\) is 0.183. b. Calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) when the volume of the cylinder is increased until the total pressure is \(1.0 \mathrm{~atm}\):

05

Write the reaction quotient (Q) and the conditions for the new equilibrium

We can write the reaction quotient, Q, for the increased volume condition as follows: \[Q=\frac{[NO_{2}]^2}{[N2O4]}\] Notice that since the total pressure is decreased to 1 atm, the reaction will shift to maintain equilibrium. In this case, the reaction shifts towards the side where the number of moles of gas is greater, which is the dissociation of N2O4. The reaction will proceed until Q = Kp.

06

Calculate the new partial pressures of the gases at equilibrium

Let 'x' be the moles of N2O4 that dissociate at the new equilibrium. Then, the moles of N2O4 will decrease by x and the moles of NO2 will increase by 2x. Given the initial partial pressures, we can calculate the new partial pressures as follows: \[P_{N2O4} = (1.26 - x) \mathrm{~atm}\] \[P_{NO2} = (0.48 + 2x) \mathrm{~atm}\]

07

Solve the equation for the new equilibrium partial pressures

Now, we can set the quotient Q equal to Kp and find the equilibrium partial pressures. From \(Q=K_p\), we have \[\frac{[(0.48 + 2x)^{2}]}{(1.26-x)}=0.183\] To solve this equation for x, we can use algebraic methods or a trial and error approach. The solution to the equation gives us x ≈ 0.09. We can now find the new equilibrium partial pressures: \[P_{N2O4} = (1.26 - 0.09) = 1.17 \mathrm{~atm}\] \[P_{NO2} = (0.48 + 2\cdot0.09) = 0.66 \mathrm{~atm}\] So, the new equilibrium pressures are \(P_{N2O4} = 1.17 \mathrm{~atm}\) and \(P_{NO2} = 0.66 \mathrm{~atm}\). c. Calculate the percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) dissociated at the new equilibrium position (total pressure = 1.00 atm):

08

Calculate the percentage of dissociated N2O4

The initial moles of N2O4 were 1.26 atm, with 0.09 moles dissociated at the new equilibrium. Therefore, the percentage dissociation at the new equilibrium position can be calculated as follows: \[\mathrm{Percentage \: dissociation} = \frac{0.09}{1.26} \times 100 = 7.14\%\] So, at the new equilibrium position (total pressure = 1.00 atm), 7.14% of the original \(\mathrm{N}_{2}\mathrm{O}_{4}(g)\) is dissociated.

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