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

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 12, Problem 102CP
Textbook Problem
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Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300. K. At equilibrium the total pressure was 110.5 torr. The reaction is

2 NO ( g ) + Br 2 ( g ) 2 NOBr ( g )

a. Calculate the value of Kp.

b. What would be the partial pressures of all species if NO and Br2, both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?

(a)

Interpretation Introduction

Interpretation: The value of the equilibrium constant is to be calculated. The partial pressure of all species is to be calculated corresponds that if initial partial pressure of both NO and Br2 is 0.30atm.

Concept introduction: The equilibrium constant (Kp) describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure.

Partial pressure of a gas is defined as the pressure applied by the each gas to the surroundings.

The equilibrium constant depends upon temperature.

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

Law of mass action states that at a given temperature the equilibrium constant is equal to the partial pressure of the products to the reactants raised the power of stoichiometric coefficient.

To determine: The value of the equilibrium constant.

Explanation of Solution

Explanation

Given

The given reaction is stated as,

2NO(g)+Br22NOBr(g)

The partial pressure of NO is 98.4torr.

The partial pressure of Br2 is 41.3torr.

The total pressure at equilibrium is 110.5torr.

The equilibrium constant of the gaseous species is calculated by the law of mass action.

For a general equation,

aA(g)+bB(g)cC(g)+dD(g)

The law of mass action is given by the formula,

kp=(PC)c(PD)d(PA)d(PB)b

Where,

  • kp is the equilibrium constant of the gas.
  • Numerator and denominator in parenthesis represent the partial pressure of the gases.
  • The power represents the stoichiometric coefficients of the reactants and the products.

In order to solve the equilibrium constant,(ICE-chart) that is the initial pressure, change in pressure and equilibrium pressure is determined.

The ICE table for the given reaction is,

NOBr2NOBrInitial98.41torr41.3torr0Change2xx+2xEquilibrium98.42x41.3x2x

For the given reaction,

2NO(g)+Br22NOBr(g)

The law of mass action is given by the formula,

kp=(PNOBr)2(PNO)2(PBr2)

Substitute the values from the ICE chart to calculate the equilibrium constant

(b)

Interpretation Introduction

Interpretation: The value of the equilibrium constant is to be calculated. The partial pressure of all species is to be calculated corresponds that if initial partial pressure of both NO and Br2 is 0.30atm.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure.

Partial pressure of a gas is defined as the pressure applied by the each gas to the surroundings.

The equilibrium constant depends upon temperature.

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

Law of mass action states that at a given temperature the equilibrium constant is equal to the partial pressure of the products to the reactants raised the power of stoichiometric coefficient.

To determine: The value of the partial pressure of all gases.

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Chapter 12 Solutions

Chemistry: An Atoms First Approach
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