Calculate the standard reaction free energy of each of the following reactions: (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), K=41\) ar \(400 \mathrm{~K}\) (b) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.0 \times 10^{4}\) at \(700 \mathrm{~K}\)

Short Answer

Expert verified
ΔG° for reaction (a) is approximately -32.619 kJ/mol, and for reaction (b) it is approximately -163.470 kJ/mol.

Step by step solution

01

Understand the Concept

The standard reaction free energy (ΔG°) can be calculated using the formula ΔG° = -RT ln(K), where R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.
02

Calculate ΔG° for Reaction (a)

Substitute the values into the formula for reaction (a). R = 8.314 J/mol·K, T = 400 K, and K = 41. The calculation will be ΔG° = - (8.314 J/mol·K) * (400 K) * ln(41).
03

Perform the Calculation for Reaction (a)

Calculate the natural logarithm of K, ln(41), and then multiply by -RT. ΔG° for reaction (a) is the product of this calculation.
04

Calculate ΔG° for Reaction (b)

Substitute the values into the formula for reaction (b). R = 8.314 J/mol·K, T = 700 K, and K = 3.0 x 10^4. The calculation will be ΔG° = - (8.314 J/mol·K) * (700 K) * ln(3.0 x 10^4).
05

Perform the Calculation for Reaction (b)

Calculate the natural logarithm of K, ln(3.0 x 10^4), and then multiply by -RT. ΔG° for reaction (b) is the product of this calculation.

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Key Concepts

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

Equilibrium Constant
When a chemical reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction, it is said to be in equilibrium. The position of the equilibrium is represented quantitatively by the equilibrium constant, denoted as K. This dimensionless value is calculated by the ratio of the products' concentrations raised to their stoichiometric coefficients to the reactants' concentrations raised to their stoichiometric coefficients, at equilibrium.

In the context of the exercise, the equilibrium constants are given for two reactions, which are pivotal in determining the favorability of the reactions under standard conditions using thermodynamics. It's important to remember that a large value of K indicates a reaction that favors the formation of products, while a small value of K indicates a reaction that favors the reactants.
Thermodynamics
Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In chemical reactions, thermodynamics focuses on energy changes, particularly the concept of free energy which predicts the spontaneity of a reaction.

A key concept in thermodynamics is that of 'spontaneity'. A spontaneous process is one that occurs without needing to be driven by an external force. The sign and magnitude of the Gibbs free energy change, ΔG, indicate if a process is spontaneous and can proceed without outside intervention. Understanding how to calculate and interpret these values is crucial for predicting the behavior of chemical reactions.
Gibbs Free Energy
Gibbs free energy, symbolized as G, is a thermodynamic quantity that represents the maximum amount of work that can be performed by a thermodynamic process at a constant temperature and pressure. The change in Gibbs free energy, denoted as ΔG, determines the spontaneity of a process. If ΔG is negative, the process is spontaneous; if positive, it is non-spontaneous. At equilibrium, ΔG is equal to zero.

The exercise demonstrates the calculation of standard reaction free energy (ΔG°) using the equation ΔG° = -RT ln(K), which elegantly ties together the equilibrium constant with thermodynamics. Here, 'R' is the universal gas constant and 'T' is the absolute temperature. A negative ΔG° value would imply that the reaction can spontaneously proceed in the forward direction under standard conditions.
Chemical Reaction Equilibrium
Chemical reaction equilibrium refers to the state of a reaction where the concentrations of reactants and products remain constant over time because the rate of the forward reaction equals the rate of the reverse reaction. This concept is a direct consequence of the law of mass action, which states that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, known as the equilibrium constant (K).

In practical terms, for the reactions in the exercise, reaching equilibrium means that the synthesis of ammonia from nitrogen and hydrogen gases and the formation of sulfur trioxide from sulfur dioxide and oxygen will proceed until a fixed ratio of reactants to products is achieved. This ratio is represented by the respective equilibrium constants. Understanding how to maneuver between the values of K and ΔG° gives insight into how changes in conditions, like temperature, can shift the position of equilibrium, impacting industrial processes like the synthesis of ammonia in Haber's process or sulfuric acid production.

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Most popular questions from this chapter

Calculate the equilibrium constant ar \(25^{\circ} \mathrm{C}\) for cach of the following reactions from data available in Appendix 2A. (a) the combustion of hydrogen: \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) the oxidation of carbon monoxide: \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})\) (c) the decompostion of limestone: \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

A mixture consisting of \(2.23 \times 10^{-3} \mathrm{~mol} \mathrm{~N}_{2}\) and \(6.69 \times 10^{-3} \mathrm{~mol} \mathrm{H}_{2}\) in a \(500-\mathrm{ml}\). container was heated to \(600 \mathrm{~K}\) and allowed to reach equilibrium. Will more ammonia be formed if that equilibrium mixture is then heated to \(700 \mathrm{~K}\) ? For \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=\) \(2 \mathrm{NH}_{3}(\mathrm{~g}), K=1.7 \times 10^{-3}\) at \(600 \mathrm{~K}\) and \(7.8 \times 10^{-5}\) at \(700 \mathrm{~K}\).

Given that \(K_{c}=61\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500 \mathrm{~K}\), calculate whether more ammonia will tend to form when a mixture of composition \(2.23 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(1.24 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(1.12 \times 10^{-4} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\) is present in a container at \(500 \mathrm{~K}\).

Consider the equilibrium \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). (a) If the partial pressure of \(\mathrm{CO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{H}_{2}\) ? (b) If the partial pressure of \(\mathrm{CO}\) is decreased, what happens to the partial pressure of \(\mathrm{CO}_{2}\) ? (c) If the concentration of \(\mathrm{CO}\) is increased, what happens to the concentration of \(\mathrm{H}_{2}\) ? (d) If the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is decreased, what happens to the equilibrium constant for the reaction?

Analysis of a reaction mixture showed that it had the composition \(0.417 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(0.524 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.122 \mathrm{~mol}-\mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 \mathrm{~K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g}) \cdot(\mathrm{a}) \mathrm{Calculare}\) the reaction quotient. (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

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