Chapter 13, Problem 50E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia (NH3). At 200°C in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium, the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the Kp value for this reaction.

Interpretation Introduction

Interpretation: Nitrogen gas and hydrogen gas at given partial pressure values are mixed. The total equilibrium pressure is given. The partial pressure of hydrogen gas at equilibrium and value of the equilibrium constant (Kp) for the given reaction 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 hydrogen gas at equilibrium and value of the equilibrium constant (Kp) for the given reaction.

The partial pressure of hydrogen gas at equilibrium for the given reaction.

Explanation

Given

The initial partial pressure of N2(g) is 1.00atm .

The initial partial pressure of H2(g) is 2.00atm .

The total pressure at equilibrium is 2.00atm .

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

N2(g)+3H2(g)2NH3(g)

The equilibrium partial pressure values are represented as,

N2(g)+3H2(g)2NH3(g)Initial(atm)1.002.000Change(atm)-x-3x+2xEquilibrium(atm)1.00-x2.00-3x2x

The total pressure at equilibrium is given to be 2.00atm .

Therefore,

PN2+PH2+PNH3=2.00atm

Where,

• PN2 is the equilibrium partial pressure of N2 .
• PH2 is the equilibrium partial pressure of H2 .
• PNH3 is the equilibrium partial pressure of NH3 .

Substitute the values of PN2 , PH2 and PNH3 in the above expression (using the ICE table).

PN2+PH2+PNH3=2.00atm(1.00x)+(2.003x)+2x=2.00atmx=0.500

The equilibrium partial pressure of N2 is calculated by the formula,

EquilibriumpartialpressureofN2=(1.00x)

Substitute the calculated value of x in the above expression.

EquilibriumpartialpressureofN2=(1.000.500)atm=0.500atm

The equilibrium partial pressure of H2 is calculated by the formula,

EquilibriumpartialpressureofH2=(2

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