   Chapter 13, Problem 96CWP

Chapter
Section
Textbook Problem

# For the reaction N 2 O 4 ( g ) ⇌ 2 NO 2 ( g ) , K p = 0.25 at a certain temperature. If 0.040 atm of N2O4 is reacted initially, calculate the equilibrium partial pressures of NO2(g) and N2O4(g).

Interpretation Introduction

Interpretation: The equilibrium partial pressure of NO2(g) and N2O4(g) is to be calculated.

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

The equilibrium constant depends upon temperature.

Law of mass action is applicable on the equilibrium reactions.

To determine: The values of equilibrium partial pressure of NO2(g) and N2O4(g) .

Explanation

Explanation

Given

The reaction is given as,

N2O4(g)2NO2(g)

The initial moles of N2O4(g) is 0.040mol .

The equilibrium constant is 0.25 .

The ICE-table in terms of partial pressure for the reaction is given as,

N2O4(g)2NO2(g)Initial0.0400Change-x+2xEquilibrium0.040-x2x

The equilibrium constant in terms of partial pressure for the above reaction is calculated by the formula,

Kp=PNO22PN2O4

Where,

• Kp is the equilibrium constant.
• PNO2 is the equilibrium partial pressure of NO2(g)

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