Need help with Part D thru F only (ignore part A - C) Exercise 4.8 An important

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Need help with Part D thru F only (ignore part A - C)

Exercise 4.8

An important metabolic step is the conversion of fumarate to malate. In aqueous solution, an enzyme (fumarase) allows equilibrium to be attained:

fumarate+H2Omalate

at 25C; the equilibrium constant K = (aM/aF) = 4.0. The activity of malate is aM, and the activity of fumarate is aF, defined on the molarity concentration scale (a = c in dilute solution).

Part A

What is the standard Gibbs free-energy change for the reaction at 25C?

Express your answer using three significant figures and include the appropriate units.

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Significant Figures Feedback: Your answer 3435Jmol(=-3.435 kJ/mol) was either rounded differently or used a different number of significant figures than required for this part.

Part B

What is the Gibbs free-energy change for the reaction at equilibrium?

Express your answer using three significant figures and include the appropriate units.

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Part C

What is the Gibbs free-energy change when 1 mol of 0.100 M fumarate is converted to 1 mol of 0.100 M malate?

Express your answer using three significant figures and include the appropriate units.

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Part D

What is the Gibbs free-energy change when 7 mol of 0.100 M fumarate are converted to 7 mol of 0.100 M malate?

Express your answer using three significant figures and include the appropriate units.

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Part E

If K = 8.0 at 35C, calculate the standard enthalpy change for the reaction; assume that the enthalpy is independent of temperature.

Express your answer using three significant figures and include the appropriate units.

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Part F

Calculate the standard entropy change for the reaction; assume that rS is independent of temperature.

Express your answer using three significant figures and include the appropriate units.

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Explanation / Answer

Part D : we have 7 mols of fumarate going to to 7 mols of malate

So moles of fumarate = moles of malate = 1 : 1

gibbs free energy change = -3.43 kJ

Part E : standard enthalpy of formation for the reaction = enthalpy of products - enthalpy of reactants

= (-5) - (-285.8 + 17.6) = -263.2 kJ/mol

Part F : at 35 oC,

standard entropy of reaction = deltaH/T = -263.2/(35+273) = -0.854 kJ/K.mol

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