   Chapter 13, Problem 53E

Chapter
Section
Textbook Problem

# At 2200°C, Kp = 0.050 for the reaction N 2 ( g ) + O 2 ( g ) ⇌ 2 NO ( g ) What is the partial pressure of NO in equilibrium with N2 and O2 that were placed in a flask at initial pressures of 0.80 and 0.20 atm, respectively?

Interpretation Introduction

Interpretation: The equilibrium constant value for the reaction between N2 and O2 , placed in a flask at an initial pressure of 0.80 and 0.20atm , is given. The partial pressure of NO is to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of pressure, it is represented as Kp .

To determine: The partial pressure of NO .

Explanation

Given

The initial partial pressure of N2(g) is 0.80atm .

The initial partial pressure of O2(g) is 0.20atm .

The value of the equilibrium constant (Kp) is 0.050 .

The reaction that takes place between hydrogen gas and nitrogen gas is,

N2(g)+O2(g)2NO(g)

At equilibrium, the equilibrium ratio is expressed by the formula,

Kp=PartialpressureofproductsPartialpressureofreactants

Where,

• Kp is the equilibrium constant in terms of partial pressure.

The equilibrium ratio for the given reaction is,

Kp=(PNO)2(PN2)(PO2) (1)

The equilibrium partial pressure values are represented as,

N2(g)+O2(g)2NO(g)Initial(atm)0.800.200Change(atm)-x-x+2xEquilibrium(atm)0.80-x0.20-x2x

According to the ICE table formed,

The equilibrium partial pressure of N2 (PN2) is 0.80x .

The equilibrium partial pressure of O2 (PO2) is 0.20x .

The equilibrium partial pressure of NO (PNO) is 2x .

Substitute the values of PN2 , PH2 and PNH3 in equation (1).

Kp=(2x)2(0.80x)(0.20x)

Substitute the given value of Kp in the above expression

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