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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

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)

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

Explanation

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

(b)

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

(c)

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

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