Question

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

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

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

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.**

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

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

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)?**

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.

Key concepts

Chemical Equilibrium

In the world of chemistry, **chemical equilibrium** is like a balance scale that doesn't tip to one side. It occurs when the rates of the forward and reverse chemical reactions are equal, meaning the concentrations of the reactants and products remain constant over time. Chemical equilibrium doesn't imply that the amounts of reactants and products are equal, rather it means they are constant.
The concept is pivotal to understanding how reactions work in practice as it relates directly to the equilibrium constant ( K_ ext{p} ), which indicates the position of the equilibrium for reactions involving gases.
In our example involving the dissociation of N_2O_4 to NO_2 , the equilibrium represents a state where the rate of N_2O_4 dissociating into NO_2 is equal to the rate of NO_2 recombining to form N_2O_4 . This dynamic balance showcases how chemical systems adjust themselves to changes in conditions like pressure or concentration.

Partial Pressure

**Partial pressure** is like the individual contribution of each gas in a mixture to the total pressure. Imagine it as each gas having its own mini pressure gauge within the larger mixture.
When we discuss gases like N_2O_4 and NO_2 in the context of chemical equilibrium, partial pressure becomes crucial for calculating the equilibrium constant ( K_ ext{p} ).
For example, if the total pressure in a container is given, you can use mole fractions to determine the partial pressures of each gas component. In our scenario, knowing that 16% of N_2O_4 has dissociated allows us to deduce the partial pressures of both N_2O_4 and NO_2 from the total pressure. This knowledge then feeds into calculating important values like K_ ext{p} , which helps us understand the behavior of the gases under equilibrium conditions.

Gas Dissociation

**Gas dissociation** refers to the process where a compound, often a molecule, breaks down into smaller molecules or atoms. Typically, this happens when energy in the form of heat is added, but not always. In the given exercise, N_2O_4 (g) undergoes dissociation to form NO_2 (g).
This reaction is a classic example of a reversible reaction, where N_2O_4 dissociates into NO_2 and NO_2 can recombine to form N_2O_4 as well.
The degree to which the gas dissociates can be affected by factors such as pressure, temperature, and concentration changes. Here, an increase in the tank volume (which lowers the total pressure) led to a change in the percentage dissociation of N_2O_4 , illustrating the delicate balance of chemical equilibrium and how it's directly influenced by these variable conditions.

Reaction Quotient

The **reaction quotient** ( Q_ ext{c} or Q_ ext{p} for gases) is a snapshot of the reaction at a given moment in time. It looks similar to the equilibrium constant ( K ext{c} or K ext{p}), as both are calculated using the same formula involving concentrations or pressures of reactants and products, but it tells us where the reaction is at a particular moment relative to equilibrium.
If Q = K , the system is at equilibrium. If Q < K , the reaction will proceed to the right, forming more products to reach equilibrium. Conversely, if Q > K , the reaction will shift to the left, forming more reactants.
In our exercise, the Q ext{p} helps us understand how the reaction adjusts when the total pressure is changed from 1.5 atm to 1.0 atm. We used the known K_ ext{p} value to solve for the new conditions at equilibrium, showing the practical applicability of understanding and using the reaction quotient.