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Chapter 13: Problem 100

Nitric oxide and bromine at initial partial pressures of \(98.4\) and \(41.3\) torr, respectively, were allowed to react at \(300 . \mathrm{K}\). At equilibrium the total pressure was \(110.5\) torr. The reaction is $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), both at an initial partial pressure of \(0.30 \mathrm{~atm}\), were allowed to come to equilibrium at this temperature?

Short Answer

Expert verified
a) For the given reaction, the value of \(K_p\) is approximately \(0.412\). b) When the initial partial pressures of NO and Br2 are both 0.30 atm, the equilibrium partial pressures are: P(NO) 鈮� 0.012 atm, P(Br2) 鈮� 0.056 atm, P(NOBr) 鈮� 0.49 atm.

Step by step solution

01

a) Calculate the value of Kp.

First, let's determine the change in partial pressures for all species. We are given the following initial pressures: P(NO)initial = 98.4 torr P(Br2)initial = 41.3 torr P(NOBr)initial = 0 torr (since NOBr initially is not present) The total equilibrium pressure is given as 110.5 torr. Let x represent the change in partial pressures: P(NO) = 98.4 - 2x torr P(Br2) = 41.3 - x torr P(NOBr) = 2x torr Now we can write the equilibrium expression: \( K_p = \frac{P(NOBr)^2}{P(NO)^2 \times P(Br_2)} \) Substitute the expressions in terms of x: \( K_p = \frac{(2x)^2}{(98.4-2x)^2 \times (41.3-x)} \) Since we have the total equilibrium pressure, we can write the equation: Total pressure = P(NO) + P(Br2) + P(NOBr) 110.5 = (98.4 - 2x) + (41.3 - x) + (2x) Solving for x: 110.5 = 98.4 + 41.3 - x x = 29.2 torr Now, substitute the value of x back into the expressions for the partial pressures at equilibrium: P(NO) = 98.4 - 2(29.2) = 40 torr P(Br2) = 41.3 - 29.2 = 12.1 torr P(NOBr) = 2(29.2) = 58.4 torr Plug the equilibrium partial pressures back into the equilibrium expression: \( K_p = \frac{58.4^2}{40^2 \times 12.1} \) Now, calculate the value of Kp: \( K_p \approx 0.412 \)
02

b) Find the equilibrium partial pressures.

Given the initial partial pressures of NO and Br2 (both at 0.30 atm), we need to find the equilibrium partial pressures for all species. First, convert the initial pressures to torr: P(NO)initial = 0.30 atm 脳 (760 torr/atm) = 228 torr P(Br2)initial = 0.30 atm 脳 (760 torr/atm) = 228 torr Let y represent the change in partial pressures: P(NO) = 228 - 2y torr P(Br2) = 228 - y torr P(NOBr) = 2y torr Using the previously calculated Kp value (0.412), we can write the equilibrium expression: \( 0.412 = \frac{(2y)^2}{(228-2y)^2 \times (228-y)} \) Solving this equation for y is a bit more challenging, so it's recommended to use a numerical solving method such as the Newton-Raphson method or trial and error. After solving for y, we get: y 鈮� 185.21 torr Now, substitute the value of y back into the expressions for the partial pressures and convert them back to atm: P(NO) = (228 - 2(185.21)) / 760 鈮� 0.012 atm P(Br2) = (228 - 185.21) / 760 鈮� 0.056 atm P(NOBr) = 2(185.21) / 760 鈮� 0.49 atm So, at equilibrium, the partial pressures of the species are: P(NO) 鈮� 0.012 atm P(Br2) 鈮� 0.056 atm P(NOBr) 鈮� 0.49 atm

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure
In gases, each component exerts its own pressure independently of the others. This is called the partial pressure. It helps to understand how much a specific gas contributes to the total pressure in a mixture. For example, if we have a mixture of nitric oxide (NO) and bromine (Br\(_2\)), we can calculate each one's contribution to the overall pressure.

The partial pressure of a gas is determined by multiplying the mole fraction of the gas by the total pressure. This concept is key in chemical equilibrium calculations involving gases.

Using the initial pressures and the total equilibrium pressures given, we can determine changes in partial pressures during a reaction. This information is used to find the equilibrium constant, a vital step in explaining how reactions reach balance.
Chemical Equilibrium
Chemical equilibrium is a state where the forward and reverse reactions occur at the same rate, and the concentrations of reactants and products remain constant. This doesn't mean the reactants and products are equal in concentration, just that their rates of formation are balanced.

In the reaction between NO and Br\(_2\), equilibrium was reached with specific pressures for each gas. Understanding equilibrium helps predict how a change in conditions (like pressure or temperature) can shift the balance of a chemical reaction.

At equilibrium, we can use these constant concentrations to calculate the equilibrium constant (K_p). This value tells us the ratio of product concentrations to reactant concentrations at equilibrium and indicates the extent to which a reaction occurs.
Reaction Quotient
The reaction quotient ( Q) is a ratio similar to the equilibrium constant, but it's calculated using the current concentrations or pressures, not those at equilibrium. It provides a snapshot of a reaction's status.

By comparing Q to K_p (the equilibrium constant), we can predict the reaction's direction:
  • If Q = K_p, the system is at equilibrium.
  • If Q < K_p, the reaction will proceed forward to form more products.
  • If Q > K_p, the reaction will move in reverse to form more reactants.
In our exercise, Q was used to determine the course of the reaction and make sure it progresses towards equilibrium, using partial pressures as a guide.

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Most popular questions from this chapter

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ The initial density of the system was recorded as \(4.495 \mathrm{~g} / \mathrm{L}\). After equilibrium was reached, the density was noted to be \(4.086 \mathrm{~g} / \mathrm{L}\). a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.

Le Ch芒telier's principle is stated (Section \(13.7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

In a study of the reaction $$ 3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) $$ at \(1200 \mathrm{~K}\) it was observed that when the equilibrium partial pressure of water vapor is \(15.0\) torr, that total pressure at equilibrium is \(36.3\) torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at \(1200 \mathrm{~K}\). (Hint: Apply Dalton's law of partial pressures.)

At a particular temperature, a 3.0-L flask contains \(2.4 \mathrm{~mol} \mathrm{Cl}_{2}\), \(1.0 \mathrm{~mol} \mathrm{NOCl}\), and \(4.5 \times 10^{-3} \mathrm{~mol}\) NO. Calculate \(K\) at this temperature for the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$

A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{~K}\), where the following reactions occurred: $$ \begin{aligned} 2 \mathrm{FeSO}_{4}(s) & \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) & \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{aligned} $$ After equilibrium was reached, the total pressure was \(0.836\) atm and the partial pressure of oxygen was \(0.0275\) atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.
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