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Chapter 24: Problem 99

The standard free energy of formation of solid maltose is $-1334.42 \mathrm{~kJ} / \mathrm{mol},\( whereas that of solid glucose is \)-910.56 \mathrm{~kJ} / \mathrm{mol}\(. What is \)\Delta G^{\circ}$ for the condensation of two glucose molecules to form maltose?

Short Answer

Expert verified
The standard free energy change for the condensation of two glucose molecules to form a maltose molecule is \(\Delta G^{\circ} = 486.70 \mathrm{~kJ/mol}\).

Step by step solution

01

Write the chemical equation for the condensation reaction.

The condensation reaction of two glucose molecules to form a maltose molecule can be represented by the following chemical equation: 2 C6H12O6 (s) → C12H22O11 (s) + H2O (l) This shows that two glucose molecules (C6H12O6) combine to form one maltose molecule (C12H22O11) and one water molecule (H2O).
02

Write the equation for calculating the standard free energy change (∆G°) for the reaction.

The standard free energy change (∆G°) for the condensation reaction can be calculated using the following equation: ∆G°(reaction) = ∆G°(products) - ∆G°(reactants) For this equation, we need to consider the free energy of formation for maltose and glucose molecules.
03

Substitute the given values of the standard free energy of formation for maltose and glucose.

We are given the following values for the standard free energy of formation: - Maltose: ∆G°(maltose) = -1334.42 kJ/mol - Glucose: ∆G°(glucose) = -910.56 kJ/mol Now we will substitute these values into the equation from Step 2: ∆G°(reaction) = ∆G°(maltose) - 2 * ∆G°(glucose)
04

Calculate the standard free energy change for the condensation reaction.

Using the information from Step 3, we can calculate the standard free energy change for the condensation reaction: ∆G°(reaction) = (-1334.42 kJ/mol) - 2 * (-910.56 kJ/mol) ∆G°(reaction) = (-1334.42 kJ/mol) + 1821.12 kJ/mol ∆G°(reaction) = 486.70 kJ/mol Thus, the standard free energy change for the condensation of two glucose molecules to form a maltose molecule is 486.70 kJ/mol.

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Most popular questions from this chapter

(a) Give the empirical formula and structural formula for a cyclic ether containing four carbon atoms in the ring. (b) Write the structural formula for a straight-chain compound that is a structural isomer of your answer to part (a).

Give the molecular formula of a hydrocarbon containing six carbon atoms that is (a) a cyclic alkane, (b) a cyclic alkene, \((\mathbf{c})\) a linear alkyne, \((\mathbf{d})\) an aromatic hydrocarbon.

(a) One test for the presence of an alkene is to add a small amount of bromine, which is a red-brown liquid, and look for the disappearance of the red-brown color. This test does not work for detecting the presence of an aromatic hydrocarbon. Explain. (b) Write a series of reactions leading to para-bromoethylbenzene, beginning with benzene and using other reagents as needed. What isomeric side products might also be formed?

(a) Can \(\mathrm{C}_{6} \mathrm{H}_{12}\) be a saturated hydrocarbon? (b) Are all molecules containing a benzene ring unsaturated?

Is 2-butanol chiral?
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