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What is the relationship between the standard free-energy change, \(\Delta G^{\circ}\), for a reaction and the equilibrium constant, \(K ?\) What is the sign of \(\Delta G^{\circ}\) when: (a) \(K>1\) (b) \(K=1\) (c) \(K<1\)

Short Answer

Expert verified
(a) \( \Delta G^{\circ} < 0 \); (b) \( \Delta G^{\circ} = 0 \); (c) \( \Delta G^{\circ} > 0 \).

Step by step solution

01

Understanding the Relationship

The relationship between the standard free-energy change, \( \Delta G^{\circ} \), and the equilibrium constant, \( K \), for a reaction at a given temperature is expressed by the equation: \[ \Delta G^{\circ} = -RT \ln K \] where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin. This equation indicates that the sign and magnitude of \( \Delta G^{\circ} \) are related to the value of \( K \).
02

Analyzing the Case: K > 1

When \( K > 1 \), the natural logarithm, \( \ln K \), is positive, causing \( -RT \ln K \) to be negative. Hence, \( \Delta G^{\circ} < 0 \). This means the reaction is spontaneous in the forward direction under standard conditions.
03

Analyzing the Case: K = 1

When \( K = 1 \), the natural logarithm, \( \ln K \), equals \( 0 \). Therefore, \( \Delta G^{\circ} = 0 \). This means the reaction is at equilibrium under standard conditions, and there is no net change in free energy.
04

Analyzing the Case: K < 1

When \( K < 1 \), the natural logarithm, \( \ln K \), is negative, causing \( -RT \ln K \) to be positive. Thus, \( \Delta G^{\circ} > 0 \). This means the reaction is non-spontaneous in the forward direction under standard conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Energy
The concept of free energy, particularly Gibbs free energy \( \Delta G^{\circ} \), is a cornerstone in thermodynamics. It helps us determine whether a chemical reaction can proceed on its own—known as spontaneity.Free energy is essentially the energy that can be converted into work at constant temperature and pressure, which is critical in chemical and biological processes.

The equation linking free energy to the equilibrium constant is: \[ \Delta G^{\circ} = -RT \ln K \]where:
  • \(R\) is the universal gas constant
  • \(T\) is the temperature in Kelvin
  • \(K\) is the equilibrium constant
This equation shows the intimate connection between the free energy change and how a reaction is skewed towards products or reactants.By observing the sign of \( \Delta G^{\circ} \), we can quickly gauge the direction of a reaction:
  • Negative \( \Delta G^{\circ} \): Reaction is spontaneous
  • Zero \( \Delta G^{\circ} \): Reaction is at equilibrium
  • Positive \( \Delta G^{\circ} \): Reaction is non-spontaneous
Equilibrium Constant
The equilibrium constant, \( K \), represents the ratio of concentrations of products to reactants at equilibrium for a given reaction.

It's a measure of the extent to which a reaction proceeds and helps us understand the balance between forward and reverse reactions.
Though its value doesn't concern time, it tells us about the position of equilibrium:
  • \( K > 1 \): Products are favored over reactants
  • \( K = 1 \): No preference; equivalent amounts of products and reactants
  • \( K < 1 \): Reactants are favored over products
The relation between \( \Delta G^{\circ} \) and \( K \) using the equation \( \Delta G^{\circ} = -RT \ln K \) can further illuminate these states. For example, a large \( K\) results in a negative \( \Delta G^{\circ} \), showing spontaneous reactions towards products, while a small \( K\) suggests positive \( \Delta G^{\circ} \), indicating non-spontaneous reactions.
Spontaneity
Spontaneity refers to the ability of a chemical reaction to occur without any external energy input. It doesn't necessarily mean how fast the reaction will take place but whether it can happen on its own.

The sign of \( \Delta G^{\circ} \, \) gives us a straightforward understanding of spontaneity:For a spontaneous reaction:
  • \( \Delta G^{\circ} < 0 \): Reaction will proceed on its own under standard conditions.
    This typically aligns with \( K > 1 \), suggesting that products are favored.
  • \( K = 1 \): Reactants and products are balanced, and \( \Delta G^{\circ} \) equals zero,
    indicating no net energy change—equilibrium.
  • \( K < 1 \): \( \Delta G^{\circ} > 0 \) shows that the reaction is non-spontaneous,
    as a result, external energy would be required for the reaction to proceed.
Understanding the connection between free energy, equilibrium constants, and spontaneity allows us to predict and control chemical processes efficiently.

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Most popular questions from this chapter

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refining, and hundreds of other processes. An intermediate step in the industrial process for the synthesis of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is the catalytic oxidation of sulfur dioxide: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad \Delta G^{\circ}=-141.8 \mathrm{~kJ}$$ Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\), given the following sets of partial pressures: (a) \(100 \mathrm{~atm} \mathrm{SO}_{2}, 100 \mathrm{~atm} \mathrm{O}_{2}, 1.0 \mathrm{~atm} \mathrm{SO}_{3}\) (b) \(2.0 \mathrm{~atm} \mathrm{SO}_{2}, 1.0 \mathrm{~atm} \mathrm{O}_{2}, 10 \mathrm{~atm} \mathrm{SO}_{3}\) (c) Each reactant and product at a partial pressure of \(1.0 \mathrm{~atm}\)

Elemental sulfur is formed by the reaction of zinc sulfide with oxygen: $$2 \mathrm{ZnS}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{~S}(s)$$ (a) If \(\Delta H^{\circ}=-289.0 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S^{\circ}=-169.4 \mathrm{~J} / \mathrm{K}\), what is \(\Delta S_{\text {total }}\) for this reaction? Is the reaction spontaneous under standard-state conditions at \(25^{\circ} \mathrm{C} ?\) (b) At what temperature, if any, will the reaction become nonspontaneous?

Indicate whether the following processes are spontaneous or nonspontaneous: (a) Heat transfer from a block of ice to a room maintained at \(25^{\circ} \mathrm{C}\) (b) Evaporation of water from an open beaker (c) Conversion of iron(III) oxide to iron metal and oxygen (d) Uphill motion of an automobile

Consider the decomposition of gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) : $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=+55.3 \mathrm{~kJ} ; \Delta S^{\circ}=+175.7 \mathrm{~J} / \mathrm{K}$$ (a) Is this reaction spontaneous under standard-state conditions at \(25^{\circ} \mathrm{C} ?\) (b) Estimate the temperature at which the reaction becomes spontaneous.

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\), an important nitrogen fertilizer, is produced industrially by the reaction $$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$ Given that \(\Delta G^{\circ}=-13.6 \mathrm{~kJ}\), calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the following sets of conditions: (a) \(10 \mathrm{~atm} \mathrm{NH}_{3}, 10 \mathrm{~atm} \mathrm{CO}_{2}, 1.0 \mathrm{M} \mathrm{NH}_{2} \mathrm{CONH}_{2}\) (b) \(0.10 \mathrm{~atm} \mathrm{NH}_{3}, 0.10 \mathrm{~atm} \mathrm{CO}_{2}, 1.0 \mathrm{M} \mathrm{NH}_{2} \mathrm{CONH}_{2}\) (c) Is the reaction spontaneous for the conditions in part (a) and/or part (b)?

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