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Chapter 17: Problem 21

Explain the difference between \(\Delta G\) and \(\Delta G^{\circ} .\)

Short Answer

Expert verified
The difference between \(\Delta G\) and \(\Delta G^{\circ}\) lies in the conditions under which they are calculated. While \(\Delta G\) is derived under the actual conditions of a reaction, \(\Delta G^{\circ}\) is calculated under 'standard' conditions (usually 298.15 K, 1 bar pressure, and specified concentrations or activities). Therefore, \(\Delta G\) provides a representation of a specific reaction's conditions, whereas \(\Delta G^{\circ}\) facilitates comparisons among reactions under standard conditions.

Step by step solution

01

Definition of \(\Delta G\)

The first term to understand is \(\Delta G\), also known as the Gibbs Free Energy change. This is the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure. It is useful for understanding whether or not a reaction will occur spontaneously.
02

Definition of \(\Delta G^{\circ}\)

\(\Delta G^{\circ}\) refers to the Standard Gibbs Free Energy Change. It is calculated under the 'standard' conditions: at 1 bar pressure, any specified concentration or activity, and a common temperature - usually 298.15 K (25 degrees Celsius). The standard state does not necessarily have to involve these conditions, however, they are the most common.
03

Difference Between \(\Delta G\) and \(\Delta G^{\circ}\)

The main difference between \(\Delta G\) and \(\Delta G^{\circ}\) is the conditions under which they are calculated. While \(\Delta G\) is not specific to any set conditions, \(\Delta G^{\circ}\) is calculated under standard conditions. This means that \(\Delta G\) is more flexible and can reflect the conditions of a specific reaction, whereas \(\Delta G^{\circ}\) provides a benchmark for reactions under 'standard' conditions.

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Most popular questions from this chapter

Describe two ways that you could measure \(\Delta G^{\circ}\) of a reaction.

In the metabolism of glucose, the first step is the conversion of glucose to glucose 6 -phosphate: glucose \(+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) $$ \Delta G^{\circ}=13.4 \mathrm{~kJ} / \mathrm{mol} $$ Because \(\Delta G^{\circ}\) is positive, this reaction does not favor the formation of products. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction and estimate the equilibrium constant for the coupled process.

The molar heat of vaporization of ethanol is 39.3 kJ/mol and the boiling point of ethanol is \(78.3^{\circ} \mathrm{C}\). Calculate \(\Delta S\) for the vaporization of 0.50 mol ethanol.

Predict whether the entropy change is positive or negative for each of these reactions: (a) \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{O}(g)+\mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

State the second law of thermodynamics in words and express it mathematically.
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