BIOCHEMISTRY 2 TERM ACCESS
BIOCHEMISTRY 2 TERM ACCESS
9th Edition
ISBN: 9781319402877
Author: BERG
Publisher: MAC HIGHER
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Chapter 8, Problem 18P
Interpretation Introduction

Interpretation:

The matching of the given Keq' values with the appropriate ΔG°' values is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by Keq. The equilibrium constant is independent of the initial amount of the reactant and product.

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Answer to Problem 18P

The correct matching of the given Keq' values with the appropriate ΔG°' values is given in the table below.

Keq' ΔG°'(kJmol1)
(a) 1 0
(b) 105 28.53kJmol1
(c) 104 22.84kJmol1
(d) 102 11.42kJmol1
(e) 101 5.69kJmol1

Explanation of Solution

(a) The given value of Keq' is 1.

The standard temperature is 25°C.

The conversion of degrees Celsius into Kelvin is done as,

0°C=273K

Thus, the given temperature becomes,

25°C=25+273K=298K

The value of universal gas constant is 8.315×103kJmol1K1.

The value of ΔG°' is calculated by the expression,

ΔG°'=RTlnKeq  (1)

Where,

  • R is the universal gas constant.
  • T is the temperature.
  • Keq is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

ΔG°'=8.315×103kJmol1K1×298K×ln(1)=8.315×103kJmol1K1×298K×0=0

Thus, the value of Gibbs free energy for Keq' equals to 1 is 0.

(b) The given value of Keq' is 105.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

ΔG°'=8.315×103kJmol1K1×298K×ln( 10 5)=8.315×103kJmol1K1×298K×11.5129=28.53kJmol1

Thus, the value of Gibbs free energy for Keq' equals to 105 is 28.53kJmol1.

(c) The given value of Keq' is 104.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

ΔG°'=8.315×103kJmol1K1×298K×ln( 104)=8.315×103kJmol1K1×298K×9.21034=22.84kJmol1

Thus, the value of Gibbs free energy for Keq' equals to 104 is 22.84kJmol1.

(d) The given value of Keq' is 102.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

ΔG°'=8.315×103kJmol1K1×298K×ln( 102)=8.315×103kJmol1K1×298K×4.6052=11.42kJmol1

Thus, the value of Gibbs free energy for Keq' equals to 102 is 11.42kJmol1.

(e) The given value of Keq' is 101.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

ΔG°'=8.315×103kJmol1K1×298K×ln( 10 1)=8.315×103kJmol1K1×298K×2.303=5.69kJmol1

Thus, the value of Gibbs free energy for Keq' equals to 101 is 5.69kJmol1.

Conclusion

Therefore, the correct matching of the given Keq' values with the appropriate ΔG°' values is done using the following relation:

ΔG°'=RTlnKeq

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