   # At 25°C. K p = 5.3 × 10 5 for the reaction N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) When a certain partial pressure of NH 3 ( g ) is put into an otherwise empty rigid vessel at 25°C, equilibrium is reached when 50.0% of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred? ### 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 12, Problem 103CP
Textbook Problem
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## At 25°C. Kp = 5.3 × 105 for the reaction N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) When a certain partial pressure of NH3(g) is put into an otherwise empty rigid vessel at 25°C, equilibrium is reached when 50.0% of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

Interpretation Introduction

Interpretation: The reaction between nitrogen gas and hydrogen gas at 25°C and the value of equilibrium constant (Kp) is given. A certain partial pressure of the ammonia formed in the reaction is put into another rigid vessel at the same temperature. The equilibrium is reached on the 50% decomposition of ammonia. The original partial pressure of ammonia 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 original partial pressure of ammonia.

The equilibrium constant expression for the reverse reaction is, Kpr=(PN2)(PH2)3(PNH3)2

### Explanation of Solution

Explanation

Given

The stated reaction is,

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

50% of the original ammonia is given to be decomposed when the reaction reaches equilibrium.

The equilibrium constant (Kp) value for the given reaction is 5.3×105.

The reverse reaction is,

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

The equilibrium constant expression for this reaction is,

Kpr=(PN2)(PH2)3(PNH3)2 (1)

Where,

• Kpr is the equilibrium constant for the reverse reaction.

The original partial pressure of ammonia is 2.1×10-3atm_

The original partial pressure of ammonia is assumed to be xatm.

The change that occurs is =50%ofx=(50100)×x=x2

The equilibrium pressures are represented as,

2NH3(g)N2(g)+3H2(g)Initial(atm)xChangex2+x4+3x4Equilibrium(atm)xx2x43x4

The equilibrium pressure of ammonia (PNH3) is (xx2)M

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