Question

Part A Calculate the standard enthalpy change for the reaction 2A+B⇌2C+2D where the heats of formation are given in the following table: Substance ΔH∘f (kJ/mol) A -241 B -391 C 197 D -497 Express your answer in kilojoules. Hints ΔH∘rxn = 273 kJ SubmitMy AnswersGive Up Correct Part B For the reaction given in Part A, how much heat is absorbed when 2.80 mol of A reacts? Express your answer numerically in kilojoules. Hints kJ SubmitMy AnswersGive Up Part C For the reaction given in Part A, ΔS∘rxn is 45.0 J/K . What is the standard Gibbs free energy of the reaction, ΔG∘rxn? Express your answer numerically in kilojoules. Hints ΔG∘rxn = kJ SubmitMy AnswersGive Up

Answer 1

2A + B ==== 2C + 2D

The enthalpy change of formation is calculated with:

Enthalpy of products - Enthalpy of reactants

2C = 2 * 197 = 394 ,

2D = -497 * 2 = -994 ,

Enthalpy of products = 394 - 994 = -600

B = -391,

2A = 2 * -241 = -482

Enthalpy of reactants = -391 - 482 = -873

Enthalpy is = -600 - (-873) = 273 KJ

This is the energy absorbed by the system for every 2 moles of A

for every mole of A we can divide the energy by 2:

273 / 2 = 136.5 KJ

For the question of 2.8 moles of A, multiply the last value by 2.8

136.5 * 2.8 = 382.2 KJ

To calculate gibbs energy you have to apply the next equation

dG = dH - TdS

Gibbs = Enthalpy - Temperature * Entropy, beware units, enthalpy is KJ and entropy is joules, the best thing is to put enthalpy in joules , i will assume a standard temperature of 298 K which is 25 C

Gibbs = 273000 - 298 * 45 = 259590 Joules

Gibbs = 259.6 KJ

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