   # Consider the following system at equilibrium at 25°C: PCl 3 ( g ) + Cl ( g ) ⇌ PCl 5 ( g ) Δ G ∘ = − 92.50 KJ What will happen to the ratio of partial pressure of PCl 5 to partial pressure of PCI 3 if the temperature is raised? Explain completely. ### 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 83AE
Textbook Problem
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## Consider the following system at equilibrium at 25°C: PCl 3 ( g ) + Cl ( g ) ⇌ PCl 5 ( g )       Δ G ∘ = − 92.50   KJ What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCI3 if the temperature is raised? Explain completely.

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of PCl5(g) and its ΔG° value is given. The effect on the ratio of partial pressure of PCl5 to the partial pressure of PCl3 due to an increase in temperature is to be explained.

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

The given value of ΔG° is 92.50kJ .

The stated reaction is,

PCl3(g)+Cl2(g)PCl5(g)

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

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

The equilibrium constant expression for the given reaction is,

Kp=(PPCl5)(PPCl3)(PPCl2)

ΔG°=RTln(Kp)

Where,

• ΔG is the free energy change for a reaction at specified pressure.
• R is the gas law constant (8.3145J/Kmol)

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