   # Consider the following reaction at 800. K: N 2 ( g ) + 3 F 2 ( g ) → 2 NF 3 ( g ) An equilibrium mixture contains the following partial pressures: P N 2 = 0.21 atm , P F 2 = 0.063 atm , P NF 3 = 0.48 . Calculate ∆G° for the reaction at 800. K. ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 77E
Textbook Problem
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## Consider the following reaction at 800. K: N 2 ( g ) + 3 F 2 ( g ) → 2 NF 3 ( g ) An equilibrium mixture contains the following partial pressures: P N 2 = 0.21   atm ,   P F 2 = 0.063   atm ,   P NF 3 = 0.48 . Calculate ∆G° for the reaction at 800. K.

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of NF3 and partial pressure of N2,F2 and NF3 is given. The value of ΔG° is to be calculated at given temperature.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, then the free energy change will be,

ΔG=0Q=K

ΔG°=RTln(K)

### Explanation of Solution

Given

Partial pressure of nitrogen, PN2 , is 0.021atm .

Partial pressure of fluorine, PF2 , is 0.063atm .

Partial pressure of fluorine, PNF3 , is 0.48atm .

The stated reaction is,

N2(g)+3F2(g)2NF3(g)

If partial pressure of reactant and product is given, then the equilibrium pressure will be expressed as Kp and its expression will be written as,

Kp=(Partialpressure of product)a(Partialpressure of reactant)b

Where,

• a is the number of moles of product.
• b is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

Kp=(PNF3)2(PN2)(PF2)3

Substitute the values of PN2,PF2 and PNF3 in the above expression.

Kp=(PNF3)2(PN2)(PF2)3=(0.48)2(0.021)(0.063)3=4.4×104_

The value of Kp is 4.4×104

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