Chemistry [hardcover]
Chemistry [hardcover]
5th Edition
ISBN: 9780393264845
Author: Geoffery Davies
Publisher: NORTON
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Chapter 14, Problem 14.1VP

(a)

Interpretation Introduction

Interpretation: The mass action expression for the reaction is to be derived. The value of Kc is to be calculated.

Concept introduction: The division of the ratio of the concentration of product raised to suitable stoichiometric coefficient with the concentrations of the reactants raised to suitable stoichiometric coefficient is called equilibrium constant for a reaction.

To determine: The mass action expression for the reaction given in the graph shown.

(a)

Expert Solution
Check Mark

Answer to Problem 14.1VP

Solution

The mass action expression for the reaction given in the graph is 3A2+B22A3B .

Explanation of Solution

Explanation

The mass action expression for the reaction between A2 and B2 will be expressed with the help of Figure 1 given below.

Chemistry [hardcover], Chapter 14, Problem 14.1VP

Figure1

In Figure1 it is shown that there is a difference in the number of moles of A2 , B2 and A3B . The graph in Figure1 describes that the concentration of A2 changes from 1.0M to 0.52M whereas the concentration of B2 changes from 1.0M to 0.84M .

The time taken is assumed to be Δt .

The formula used to calculate the rate of change of concentration of A2 is ,

RateofchangeofconcentrationofA2withtime=ΔA2Δt (1)

Where,

  • ΔA2 is change in concentration of A2 .

The value of change in concentration of A2 is calculated by subtracting initial value from final value of A2 ,

ΔA2=1.0M0.52M=0.48M

Substitute the value of ΔA2 in equation (1),

RateofchangeofconcentrationofA2withtime=0.48MΔt (2)

The formula used to calculate the rate of change of concentration of B2 is ,

RateofchangeofconcentrationofB2withtime=ΔB2Δt (3)

Where,

  • ΔB2 is change in concentration of B2 .

The value of change in concentration of B2 is calculated by subtracting initial value from final value of B2 ,

ΔB2=1.0M0.84M=0.16M

Substitute the value of ΔB2 in equation (3),

RateofchangeofconcentrationofB2withtime=0.16MΔt (4)

The formula used to calculate the rate of change of concentration of A3B is,

RateofchangeofconcentrationofA3Bwithtime=ΔA3BΔt (5)

Where,

  • ΔA3B is change in concentration of A3B .

The value of change in concentration of A3B will be calculated by subtracting initial value from final value of A3B ,

ΔA3B=0.32M0.0M=0.32M

Substitute the value of A3B in equation (5).

RateofchangeofconcentrationofA3Bwithtime=0.32MΔt (6)

The ratio of rates is calculated as,

Ratioofrates=ΔA2Δt:ΔB2Δt:ΔA3BΔt

Substitute the values of ΔA2Δt , ΔB2Δt and ΔA3BΔt in the above expression,

Ratioofrates=0.48Δt:0.16Δt:0.32Δt==3:1:2

As the ratio of the rates is equal to the order of the reaction for the same value of Δt then,

[A2]:[B2]:[A3B]=3:1:2

Substitute the values of ratio as the order of the reaction that gives the reaction as,

3A2+B22A3B

Hence, the mass action expression for the reaction given in the graph is 3A2+B22A3B .

(b)

Interpretation Introduction

To determine: The value of Kc is to be calculated.

(b)

Expert Solution
Check Mark

Answer to Problem 14.1VP

Solution

The value of equilibrium constant Kc=[A3B]2[A2]3[B2]

Explanation of Solution

Explanation

The value of equilibrium constant Kc is calculated by the expression,

Kc=[Product][ReactantA][ReactantB]

Substitute the reactants and products at their respective places in the above expression,

Kc=[A3B]2[A2]3[B2]

Therefore the value of Kc=[A3B]2[A2]3[B2]

Conclusion

  1. a) The mass action expression for the reaction given in the graph is 3A2+B22A3B .
  2. b) The value of equilibrium constant Kc=[A3B]2[A2]3[B2]

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Chapter 14 Solutions

Chemistry [hardcover]

Ch. 14.7 - Prob. 11PECh. 14.7 - Prob. 12PECh. 14.8 - Prob. 13PECh. 14 - Prob. 14.1VPCh. 14 - Prob. 14.2VPCh. 14 - Prob. 14.3VPCh. 14 - Prob. 14.4VPCh. 14 - Prob. 14.5VPCh. 14 - Prob. 14.6VPCh. 14 - Prob. 14.7QPCh. 14 - Prob. 14.8QPCh. 14 - Prob. 14.9QPCh. 14 - Prob. 14.10QPCh. 14 - Prob. 14.11QPCh. 14 - Prob. 14.12QPCh. 14 - Prob. 14.13QPCh. 14 - Prob. 14.14QPCh. 14 - Prob. 14.15QPCh. 14 - Prob. 14.16QPCh. 14 - Prob. 14.17QPCh. 14 - Prob. 14.18QPCh. 14 - Prob. 14.19QPCh. 14 - Prob. 14.20QPCh. 14 - Prob. 14.21QPCh. 14 - Prob. 14.22QPCh. 14 - Prob. 14.23QPCh. 14 - Prob. 14.24QPCh. 14 - Prob. 14.25QPCh. 14 - Prob. 14.26QPCh. 14 - Prob. 14.27QPCh. 14 - Prob. 14.28QPCh. 14 - Prob. 14.29QPCh. 14 - Prob. 14.30QPCh. 14 - Prob. 14.31QPCh. 14 - Prob. 14.32QPCh. 14 - Prob. 14.33QPCh. 14 - Prob. 14.34QPCh. 14 - Prob. 14.35QPCh. 14 - Prob. 14.36QPCh. 14 - Prob. 14.37QPCh. 14 - Prob. 14.38QPCh. 14 - Prob. 14.39QPCh. 14 - Prob. 14.40QPCh. 14 - Prob. 14.41QPCh. 14 - Prob. 14.42QPCh. 14 - Prob. 14.43QPCh. 14 - Prob. 14.44QPCh. 14 - Prob. 14.45QPCh. 14 - Prob. 14.46QPCh. 14 - Prob. 14.47QPCh. 14 - Prob. 14.48QPCh. 14 - Prob. 14.49QPCh. 14 - Prob. 14.50QPCh. 14 - Prob. 14.51QPCh. 14 - Prob. 14.52QPCh. 14 - Prob. 14.53QPCh. 14 - Prob. 14.54QPCh. 14 - Prob. 14.55QPCh. 14 - Prob. 14.56QPCh. 14 - Prob. 14.57QPCh. 14 - Prob. 14.58QPCh. 14 - Prob. 14.59QPCh. 14 - Prob. 14.60QPCh. 14 - Prob. 14.61QPCh. 14 - Prob. 14.62QPCh. 14 - Prob. 14.63QPCh. 14 - Prob. 14.64QPCh. 14 - Prob. 14.65QPCh. 14 - Prob. 14.66QPCh. 14 - Prob. 14.67QPCh. 14 - Prob. 14.68QPCh. 14 - Prob. 14.69QPCh. 14 - Prob. 14.70QPCh. 14 - Prob. 14.71QPCh. 14 - Prob. 14.72QPCh. 14 - Prob. 14.73QPCh. 14 - Prob. 14.74QPCh. 14 - Prob. 14.75QPCh. 14 - Prob. 14.76QPCh. 14 - Prob. 14.77QPCh. 14 - Prob. 14.78QPCh. 14 - Prob. 14.79QPCh. 14 - Prob. 14.80QPCh. 14 - Prob. 14.81QPCh. 14 - Prob. 14.82QPCh. 14 - Prob. 14.83QPCh. 14 - Prob. 14.84QPCh. 14 - Prob. 14.85QPCh. 14 - Prob. 14.86QPCh. 14 - Prob. 14.87QPCh. 14 - Prob. 14.88QPCh. 14 - Prob. 14.89QPCh. 14 - Prob. 14.90QPCh. 14 - Prob. 14.91QPCh. 14 - Prob. 14.92QPCh. 14 - Prob. 14.93APCh. 14 - Prob. 14.94APCh. 14 - Prob. 14.95APCh. 14 - Prob. 14.96APCh. 14 - Prob. 14.97APCh. 14 - Prob. 14.98APCh. 14 - Prob. 14.99AP
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