Consider the unbalanced equation: (b) Use the data in Appendix B and ΔG°f for IO 3 - (aq)= -128.0 kJ/mol to calculate ΔG° for the reaction at 25 °C.

Hello everyone today. We have the following problem. Consider the disproportion ation of mercury chloride under UV light. Use the following data to calculate the value the value of our Gibbs. Free energy for this reaction at one atmospheres and at 25°C. So let's recall the equation or the formula for our Gibbs. Free energy for reaction. It's going to be the sum of the gibbs. Free energy of our products minus the sum of our gifts. Free energy of our reactant. So what is that gonna look like when we write it out? Well we see That we have one mole of our mercury. And when we multiply it we're going to get zero Because it's gibbs free energy is zero. We are then going to add that to our one more of our mercury chloride. But this time we can multiply this by our negative 1 78. kg joules primal. And lastly we're going to subtract because those were our products from our reactant which associated which was our mercury chloride. And we had one mole of that. And then we multiplied it by its gibbs. Free energy change Which was negative 2 10. kg per mole. When we solve for our Gibbs free energy for this reaction, we're gonna get 32 .1 as our final answer. Overall I hope this helped until next time

Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

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McMurry 8th Edition

Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

Problem 141b

Chapter 18, Problem 141b

Consider the unbalanced equation: (b) Use the data in Appendix B and ΔG°f for IO₃^-(aq)= -128.0 kJ/mol to calculate ΔG° for the reaction at 25 °C.

Verified step by step guidance

insert step 1> Identify the balanced chemical equation for the reaction.

insert step 2> Use the standard Gibbs free energy of formation (ΔG°f) values from Appendix B for each reactant and product in the balanced equation.

insert step 3> Calculate the ΔG° for the reaction using the formula: ΔG° = Σ(ΔG°f of products) - Σ(ΔG°f of reactants).

insert step 4> Substitute the given ΔG°f value for IO_3^-(aq) and the values from Appendix B into the equation.

insert step 5> Perform the arithmetic to find the ΔG° for the reaction at 25 °C.

Verified Solution

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy (ΔG)

Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It is a crucial concept in predicting the spontaneity of a reaction; a negative ΔG indicates that a reaction can occur spontaneously, while a positive ΔG suggests non-spontaneity.

Standard Gibbs Free Energy of Formation (ΔG°f)

The Standard Gibbs Free Energy of Formation (ΔG°f) is the change in Gibbs free energy when one mole of a compound is formed from its elements in their standard states. This value is essential for calculating the ΔG of a reaction, as it allows for the determination of the energy changes associated with the formation of reactants and products.

Reaction Quotient and Equilibrium

The reaction quotient (Q) is a measure of the relative concentrations of products and reactants at any point in a reaction. At equilibrium, Q equals the equilibrium constant (K). Understanding how ΔG relates to Q and K is vital for predicting the direction of a reaction and calculating the ΔG under non-standard conditions.

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Reaction Quotient Q

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Chloroform has ΔHvaporization = 29.2 kJ>mol and boils at 61.2 °C. What is the value of ΔSvaporization for chloroform?

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Consider the unbalanced equation: (a) Balance the equation for this reaction in basic solution.

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Consider the unbalanced equation: I₂(s) → I^-(aq) + IO₃^-(aq) (d) What pH is required for the reaction to be at equilibrium at 25°C when [I^-] = 0.10M and [IO₃^-] = 0.50 M?

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Methanol (CH₃OH) is made industrially in two steps from CO and H₂. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH₄):

Step 1. CO(g) + 2 H₂(g) S CH₃OH(l) ΔS° = - 332 J/K

Step 2. CH₃OH₁l₂ → CH₄(g) + 1/2 O₂(g) ΔS° = 162 J/K

(k) Calculate an overall ΔG°, ΔH°, and ΔS° for the formation of CH₄ from CO and H₂.

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Textbook Question

Methanol (CH₃OH) is made industrially in two steps from CO and H₂. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH₄):

Step 1. CO(g) + 2 H₂(g) S CH₃OH(l) ΔS° = - 332 J/K

Step 2. CH₃OH₁l₂ → CH₄(g) + 1/2 O₂(g) ΔS° = 162 J/K

(l) Is the overall reaction spontaneous at 298 K?

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Consider the unbalanced equation: (b) Use the data in Appendix B and ΔG°f for IO3-(aq)= -128.0 kJ/mol to calculate ΔG° for the reaction at 25 °C.

Verified Solution

Key Concepts

Gibbs Free Energy (ΔG)

Standard Gibbs Free Energy of Formation (ΔG°f)

Reaction Quotient and Equilibrium

Consider the unbalanced equation: (b) Use the data in Appendix B and ΔG°f for IO₃^-(aq)= -128.0 kJ/mol to calculate ΔG° for the reaction at 25 °C.