Consider the reaction: 2 NO(g) + Br₂(g) ⇌ 2 NOBr(g) K_p = 28.4 at 298K In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br₂ is 126 torr. What is the partial pressure of NOBr in this mixture?

검증된 단계별 안내

Step 1: Understand the problem. We are given the equilibrium constant (Kp) for the reaction 2NO(g) + Br2(g) ⇌ 2NOBr(g) and the partial pressures of NO and Br2. We need to find the partial pressure of NOBr.

Step 2: Write down the expression for the equilibrium constant (Kp) for the given reaction. Kp = [NOBr]^2 / ([NO]^2 * [Br2])

Step 3: Convert the given pressures from torr to atmospheres because the equilibrium constant is usually given in terms of atmospheres. 1 atm = 760 torr. So, [NO] = 108 torr / 760 torr/atm and [Br2] = 126 torr / 760 torr/atm.

Step 4: Substitute the known values into the equilibrium expression and solve for [NOBr]. Remember that [NOBr] = sqrt(Kp * [NO]^2 * [Br2]).

Step 5: The result will be the partial pressure of NOBr in atmospheres. If needed, convert it back to torr by multiplying by 760 torr/atm.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Equilibrium Constant (Kp)

The equilibrium constant (Kp) is a numerical value that expresses the ratio of the partial pressures of the products to the reactants at equilibrium for a given reaction at a specific temperature. For the reaction 2NO(g) + Br2(g) ⇌ 2NOBr(g), Kp is calculated using the formula Kp = (P_NOBr^2) / (P_NO^2 * P_Br2), where P represents the partial pressures of the gases involved.

Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture. It can be determined using Dalton's Law, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of its individual components. In this context, the partial pressures of NO and Br2 are given, and the partial pressure of NOBr needs to be calculated using the equilibrium constant.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to counteract the change and restore a new equilibrium. In this reaction, if the concentrations of NO or Br2 were to change, the system would shift to either the left or right to re-establish equilibrium, which is relevant when considering how the partial pressures relate to Kp.

Consider the reaction: 2 NO(g) + Br2(g) ⇌ 2 NOBr(g) Kp = 28.4 at 298K In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br2 is 126 torr. What is the partial pressure of NOBr in this mixture?