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Chemistry: An Atoms First Approach

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

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 12, Problem 115IP
Textbook Problem
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For the reaction

NH 3 ( g ) + H 2 S ( g ) NH 4 HS ( s )

K = 400. at 35.0°C. If 2.00 moles each of NH3, H2S, and NH4HS are placed in a 5.00-L vessel, what mass of NH4HS will be present at equilibrium? What is the pressure of H2S at equilibrium?

Interpretation Introduction

Interpretation: For the given reaction at equilibrium the mass of NH4HS is to be calculated. The pressure of H2S at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kc describes the ratio of the reactant to the product on the equilibrium conditions in terms of molar concentration.

The equilibrium constant depends upon temperature.

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

Law of mass action states that at a given temperature the equilibrium constant is equal to the partial pressure of the products to the reactants raised the power of stoichiometric coefficient.

To determine: The mass of the NH4HS and the pressure of the H2S gas at equilibrium.

Explanation of Solution

Explanation

Given

The reactions are given as,

NH3(g)+H2S(g)NH4HS(s)

At 45°C for the given reaction the equilibrium constant =400

The number of moles of the NH3 is 2 .

The number of moles of the H2S is 2 .

The number of moles of the NH4HS is 2 .

The concentration of all the species from the given data is calculated as,

C=nV

Where,

  • C is the concentration.
  • n is the number of moles.
  • V is the volume in liter.

Substitute the values of C , n and V in the above equation.

The concentration of NH3=0.4M

The concentration of H3S=0.4M

The concentration of NH4HS=0.4M

An ICE table is formed.

The change in concentration is assumed to be x .

The product NH4HS is solid. Therefore, the concentration of this is unity.

NH3(g)+H2S(g)NH4HS(s)Initialconcentration0.40.40.4ChangeinconcentrationxxEquilibriumconcentration0.4x0.4x

The equilibrium constant in terms of concentration for the given reaction is given by the formula.

K=1[NH3][H2S]

Where,

  • [NH3] is the equilibrium concentration of NH3
  • [H2S] is the equilibrium concentration of H2S

Substitute the values of the concentrations from the ICE-table in the above equation.

K=1[NH3][H2S]400=1(0.4x)(0.4x)400=1(0.4x)2x=0.35

As the product NH4HS is in solid phase so moles of it not changes during the attainment of equilibrium. Therefore, at equilibrium 2mols of NH4HS is present along with the more amount of NH4HS formed from the reaction.

The amount of NH4HS from the reactant is calculated by the formula,

Mol=C×V

Where,

  • Mol is the amount of the product.
  • C is the concentration of the reactants.
  • V is the volume of the reactants.

Substitute the values of concentration and volume, the amount of NH4HS at equilibrium is obtain as,

Mol=C×V=0.35M×5.00liter=3.75mol

The molar mass of NH4HS=51.113g/mol

Therefore, the mass of NH4HS at equilibrium is calculated by the formula,

MassofNH4HS=Numberofmoles×Molarmass=3

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Chapter 12 Solutions

Chemistry: An Atoms First Approach
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