### Determining the Standard Gibbs Free Energy Change (ΔG⁰ₓₙ)#### Problem Statement:Calculate the value of ΔG⁰ₓₙ for the reaction:\[ 2C \rightarrow 3A + 4B \]#### Given Information:For the reaction:\[C \rightarrow \frac{3}{2} A + 2B\]the standard Gibbs free energy change (ΔG⁰ₓₙ) is given as:\[ ΔG⁰ₓₙ = 240.5 \text{ kJ/mol} \]#### Explanation:To determine the ΔG⁰ₓₙ for the target reaction, consider the given reaction and how it relates to the target reaction.1. The given reaction: \[ C \rightarrow \frac{3}{2} A + 2B \] has a ΔG⁰ₓₙ value of 240.5 kJ/mol.2. For the target reaction: \[ 2C \rightarrow 3A + 4B \] Notice that the target reaction is exactly twice the given reaction: \[ 2 \times \left( C \rightarrow \frac{3}{2} A + 2B \right) \] which simplifies to: \[ 2C \rightarrow 3A + 4B \]3. Therefore, the ΔG⁰ₓₙ for the target reaction will be twice that of the given reaction: \[ ΔG⁰ₓₙ = 2 \times 240.5 \text{ kJ/mol} = 481.0 \text{ kJ/mol} \]### Conclusion:The standard Gibbs free energy change (ΔG⁰ₓₙ) for the reaction \(2C \rightarrow 3A + 4B\) is 481.0 kJ/mol.

Answered: Image /qna-images/answer/9718c1bf-fc2e-4c2d-a79b-faae29e5154f.jpg

Answered: What is the value of AG 2C 3A + 4B ?…

Science

Chemistry

What is the value of AG 2C 3A + 4B ? Given that 3/2 A + 2B rxn for the reaction: AG° = 240.5 kJ/mol rxn

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Please don't provide handwritten solution
Calculate AH for the reaction, C,H,(g) + H,(g) → C,H,(g) Given: (a) 2 C,H,(g) + 5 0,(g) →4 CO,(g) +2 H,O(1) AH =-2599.2 kJ (b) C,H,(g) + 3 0,(g) → 2 CO,(g) + 2 H,O(1) AH = -1410.9 kJ (c) H,(g) + 12 O,(g) → H,O(1) AH = -285.8 kJ O -174.5 kJ -4295.9 kJ -902.8 kJ O -2996.3 kJ
What is the standard enthalpy of the reaction, in kJ, at 298K for: 2 A (aq) + 1 B (aq) → 1 C (aq) + 2 D (aq) if ΔH° (A)= 83.2868 kJ/mol , ΔH° (B) = -8.2493 kJ/mol, ΔH° (C) = 87.6656 kJ/mol, and ΔH° (D) = -83.9036 kJ/mol
1. Using Hess's Law, calculate AH for this reaction: (T: 4) 2H3BO3(aq) →→→ B₂O3(s) + 3H₂O(l) (1) H₂BO3(aq) → HBO₂(aq) + H₂O(E) AH = -0.020 kJ (2) 2B₂O3(s) + H₂O(l) → H₂B4O7(s) AH = -17.5 kJ (3) H₂B4O7(s) + H₂O(l) → 4HBO₂(aq) AH = -11.3 kJ
1. Given the following reactions and data, use addition (similar to Hess's Law) to calculate AG° for H3PO4 (s). (Hint: The overall reaction is for the formation of 1 mol of H3PO4 (s) from elements in their natural state (elemental phosphorus is P4 (s)). PA () + 5 02@ → P,O10 (5) AG; - 2698 kJ P,O10 (9) + 6 H0 m → 4 H3P04 () H2 ( + ½ 02 (2) → H;O @ AG° = – 396 kJ - 228.6 kJ AG; : AG vap H,O 0 → H2O @ 8.6 kJ Required work: * rewrite the reactions above in the correct direction, using multipliers to change coefficients and modify AG° values if necessary * show that after cancelling, the sum of your reactions gives the desired overall reaction Write the goal equation first?
Calculate AH for the following reaction: IXI C(s) +H,0(g)→Co(g) - H, (3) Use the following reactions and given AH values: C(s) + O2 (g)→CO, (g), AH- 393.5 kJ 20(g) + 02 (g)→2CO2(g), AH = -566.0 kJ 2H2 (g) + O2 (g)→2H2O(g). AH = -483.6 kJ Express your answer using four significant figures.
The overall reaction for the corrosion of iron (rusting) by oxygen is 4Fe(s) + 3O2 --> 2Fe2O3 (s) b. Calculate the equilibrium constant using ΔGo = -RT ln K.
Consider the unbalanced equation for the neutralization of acetic acid: αHC2H3O2(aq)+βBa(OH)2(aq) →γH2O(l)+δBa(C2H3O2)2(aq) Balance the equation. Give your answer as an ordered set of numbers α, β, γ, δ. Your answers should reflect the lowest whole-number ratio of coefficients.
What is the value of the equilibrium constant for the following reaction at 25C with a delta Grxn = -6.38 kJ? H2(g) + Cl2(g) <-> 2HCl(g)
When 50.0 ml of 0.100 M Pb(NO3)2 solution at 20.00oC is added to 50.0 ml of 0.100 M KI solution also at 20.00oC, solid PbI2 is produced according to the equation below. The temperature of the mixture increases to 20.52oC. Calculate ΔH for the reaction below. The density of all solutions is 1.00 g/ml. The specific heat of the mixture is 4.184 J/g· oC Pb(NO3)2 (aq) + 2 KI (aq) → PbI2 (s) + 2 KNO3 (aq)