Chapter 10: Problem 34

The sugars glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q),\) and sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q),\) have \(\Delta G_{f}^{\prime}\) values of \(-915 \mathrm{kJ}\) and \(-1551 \mathrm{kJ}\) respectively. Is the hydrolysis reaction, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\) \(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q),\) likely to occur \(?\)

Short Answer

Expert verified

Yes, the hydrolysis reaction is likely to occur, as indicated by the negative Gibbs Free Energy \(\Delta G = -279 kJ/mol\).

Step by step solution

- Formulate the Reaction

The given reaction is Sucrose(aq) + H2O(l) \rightarrow 2 Glucose(aq). It's important to note that there are 2 glucose molecules produced in the reaction.

- Calculate Gibbs Free Energy for Reactants and Products

The \(\Delta G^{\prime}_{f}\) values given are for 1 mol of sucrose and 1 mol of glucose. For the reactants, there's 1 mol of sucrose \(\Delta G_{Reactants} = -1551 kJ/mol\). For the products, there's 2 mol of glucose. Thus, \(\Delta G_{Products} = 2 * -915 kJ/mol = -1830 kJ/mol\).

- Calculate the Change in Gibbs Free Energy

Using the formula \(\Delta G = \Delta G_{Products} - \Delta G_{Reactants}\), the change in Gibbs Free Energy will be \(\Delta G = -1830 kJ/mol - (-1551 kJ/mol) = -279 kJ/mol\).

- Determine if Reaction is Spontaneous

Based on the calculations \(\Delta G == -279 kJ/mol\) is negative. A negative \(\Delta G\) indicates that the reaction is spontaneous and likely to occur.

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Short Answer

Step by step solution

- Formulate the Reaction

- Calculate Gibbs Free Energy for Reactants and Products

- Calculate the Change in Gibbs Free Energy

- Determine if Reaction is Spontaneous

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