Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Textbook Question
Chapter 15, Problem 55GQ

The reaction of hydrogen and iodine to give hydrogen iodide has an equilibrium constant, Kc, of 56 at 435 °C.

  1. (a) What is the value of Kp?
  2. (b) Suppose you mix 0.045 mol of H2 and 0.045 mol of I2 in a 10.0-L flask at 425 °C. What is the total pressure of the mixture before and after equilibrium is achieved?
  3. (c) What is the partial pressure of each gas at equilibrium?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of KP in the reaction of hydrogen iodide formation has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure[Kp]: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

aA(g)+bB(g)cC(g)+dD(g)KP=PCc×PDdPAa×PBb

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

Kp=Kc(RT)ΔnKc-EquilibriumconstantintermsofconcentrationKp-EquilibriumconstantintermsofpressureΔn-changeinnumberofmoles

Answer to Problem 55GQ

The value of KP in the given reaction is 56

Explanation of Solution

Given:

H2(g)+I2(g)2HI(g)KC=56T=435°C=708K

Using the equation Kp=Kc(RT)Δn, we can calculate the value of KP.

Kp=Kc(RT)Δn=56×(0.082×708)0sinceΔn=2-2=0=56

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The total pressure of the mixture before and after equilibrium has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure[Kp]: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

aA(g)+bB(g)cC(g)+dD(g)KP=PCc×PDdPAa×PBb

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

Kp=Kc(RT)ΔnKc-EquilibriumconstantintermsofconcentrationKp-EquilibriumconstantintermsofpressureΔn-changeinnumberofmoles

Answer to Problem 55GQ

The total pressure of the mixture before equilibrium and after equilibrium is 0.52atm.

Explanation of Solution

Given:

No.ofmolesofH2=0.045molNo.ofmolesofI2=0.045molV=10LT=425°C=698K

From the reaction H2(g)+I2(g)2HI(g), it is clear that one mole of hydrogen and one mole of iodine combine to form two ,moles of hydrogen iodide. Therefore

0.045molH2andI2form0.09molofHI

Total pressure can be calculated using ideal gas equation

PV=nRT

P=nRTV=0.09×0.082×69810=0.52atm

Total pressure of the mixture is 0.52atm.

The total pressure before and after equilibrium will be the same since the amount of gas has not changed.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The partial pressure of each gas in the mixture has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure[Kp]: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

aA(g)+bB(g)cC(g)+dD(g)KP=PCc×PDdPAa×PBb

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

Kp=Kc(RT)ΔnKc-EquilibriumconstantintermsofconcentrationKp-EquilibriumconstantintermsofpressureΔn-changeinnumberofmoles

Answer to Problem 55GQ

The partial pressures of each gas in the mixture is PH2=PI2=0.052PHI=0.12atm

Explanation of Solution

Given that initially 0.045molofH2andI2 are present.

H2(g)+I2(g)2HI(g)Initialconcentration0.045mol0.045mol0Afterreactingxmoles0.045-x0.045-x2x

For the reaction H2(g)+I2(g)2HI(g)

Kc=[HI]2[H2][I2]56=[2x]2[0.045-x]256×[0.045-x]2=4x252x2-5.04x+0.1134=0x=0.06

PH2=PI2=0.052PHI=0.12atm

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Chapter 15 Solutions

Chemistry & Chemical Reactivity

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