Question

Calculate the standard Gibbs free energy of each of the following reactions: (a) \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}), K=6.8\) at 1200 . \(\mathrm{K}\) (b) \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) (s) \(\rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}), K=1.1 \times 10^{-12}\) at \(298 \mathrm{~K}\)

Short answer

\( \triangle G^{\theta}_{\text{(a)}} = -8.314 \times 1200 \times \text{ln} 6.8 \), \( \triangle G^{\theta}_{\text{(b)}} = -8.314 \times 298 \times \text{ln} (1.1 \times 10^{-12}) \).

Step-by-step solution

Understand the formula for Gibbs free energy change

The standard Gibbs free energy change for a reaction at a given temperature can be calculated using the equation: \( \triangle G^{\theta} = -RT \times \text{ln} K \), where \( \triangle G^{\theta} \) is the standard Gibbs free energy change, \( R \) is the gas constant (8.314 J/mol K), \( T \) is the temperature in kelvins, and \( K \) is the equilibrium constant of the reaction.

Calculate the standard Gibbs free energy change for reaction (a)

For the reaction \( \text{I}_{2}(g) \rightleftharpoons 2 \text{I}(g)\) with \( K=6.8 \) at \( 1200 \text{ K} \), plug these values into the equation. \[ \triangle G^{\theta}_{\text{(a)}} = -8.314 \times 1200 \times \text{ln} 6.8 \] Now calculate the value to find \( \triangle G^{\theta}_{\text{(a)}} \).

Calculate the standard Gibbs free energy change for reaction (b)

For the reaction \( \text{Ag}_{2}\text{CrO}_{4}(s) \rightleftharpoons 2 \text{Ag}^{+}(aq) + \text{CrO}_{4}^{2-}(aq)\) with \( K=1.1 \times 10^{-12} \) at \( 298 \text{ K} \), use the equation. \[ \triangle G^{\theta}_{\text{(b)}} = -8.314 \times 298 \times \text{ln} (1.1 \times 10^{-12}) \] Calculate the value to find \( \triangle G^{\theta}_{\text{(b)}} \).

Perform the calculations

Use a calculator or a computational tool to evaluate the natural logarithms and the multiplications for both reactions to find the numerical values of \( \triangle G^{\theta} \) for reactions (a) and (b).

Key concepts

Chemical Equilibrium

In the study of chemical reactions, chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time, but not necessarily equal. It's like a tug of war where both sides are equally strong, and no side wins.

At equilibrium, the reaction does not stop; instead, it continues with both reactions occurring at the same rate, leading to a dynamic balance. Understanding equilibrium is crucial as it helps chemists control reactions to yield the maximum amount of desired products. In relation to Gibbs free energy, when a system is at equilibrium, the changes in free energy are zero since the system is in its most stable state.

Equilibrium Constant

The equilibrium constant (K) is a number that reflects the ratio of product concentrations to reactant concentrations at chemical equilibrium, raised to the power of their stoichiometric coefficients. The value of K gives us insight into the position of equilibrium:

• If K>1, the reaction favors products, meaning it lies to the right.
• If K<1, it favors reactants, meaning it lies to the left.

A large K indicates a strong tendency toward products at equilibrium, whereas a small K suggests that reactants are more favored.

It's important to note that K is temperature-dependent and does not include the concentration of solid or liquid pure substances. For Gibbs free energy calculations, the equilibrium constant is a critical input that can predict the spontaneity of a reaction.

Thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, and energy of a system. It’s pivotal to chemistry since it allows us to predict whether a reaction is spontaneous through the concept of Gibbs free energy. A key principle is that spontaneous reactions are characterized by a decrease in the total free energy of a system. This doesn't mean that they occur rapidly, but rather that they have the potential to proceed without external energy input.

There are three main laws of thermodynamics that set rules for how energy moves and transforms: the first law deals with conservation of energy, the second law sheds light on increased entropy, or disorder, and the third law sets absolute zero as a point where entropy also reaches its minimum. Gibbs free energy ties into this beautifully because it merges enthalpy, which is heat content, and entropy, to give us a single value that can predict the spontaneity of reactions.

Natural Logarithm

The natural logarithm (denoted as ln) is a mathematical function that’s particularly significant in the realm of exponential relationships, such as those found in growth and decay processes. It corresponds to the time needed to reach a certain level of continuous growth. For Gibbs free energy calculations, the natural logarithm transforms the equilibrium constant into a component that can be used alongside other constants to determine the free energy of a system.

It's important for students to be comfortable with the use of the natural logarithm in chemistry because it often appears in equations dealing with kinetics and equilibrium. Moreover, understanding the properties of ln, such as how to manipulate it in algebraic expressions, can be invaluable for solving thermodynamic problems effectively.