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Chapter 7: Problem 77

One glucose molecule, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}),\) is converted to two lactic acid molecules, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s})\) during glycolysis. Given the combustion reactions of glucose and lactic acid, determine the standard enthalpy for glycolysis. $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s})+6 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2808 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) & \mathrm{COOH}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \\ 3 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(1) & \Delta H^{\circ}=-1344 \mathrm{kJ} \end{aligned}$$

Short Answer

Expert verified
The standard enthalpy for glycolysis is \(ΔH = -1120 kJ\) based on the given combustion reactions and the formula used to calculate the enthalpy of the glycolysis process.

Step by step solution

01

Identifying Reactions

Identify the combustion reactions of both glucose and lactic acid as provided:\n\nFor glucose: \n\n\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s})+6 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), \( \Delta H^{\circ}=-2808 \mathrm{kJ}\)\n\nFor lactic acid: \n\n\(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), \( \Delta H^{\circ}=-1344 \mathrm{kJ}\)
02

Formulate Glycolysis Reaction

Formulate the reaction for the process of glycolysis involving glucose and lactic acid. The glycolysis reaction can be written as follows: \n\n\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \rightarrow 2\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s})\)
03

Calculate Enthalpy for Glycolysis

Using Hess's Law, we can calculate the ΔH of the glycolysis process by subtracting the ΔH of glucose from twice the ΔH of lactic acid. That is, \n\nΔH (glycolysis) = 2ΔH (lactic acid) - ΔH (glucose)\n\nSubstituting the given ΔH values into this equation, we get: \n\nΔH (glycolysis) = 2*(-1344 kJ) - (-2808 kJ)\n\nSolving this will give the standard enthalpy for the process of glycolysis.

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Most popular questions from this chapter

\(\mathrm{CCl}_{4},\) an important commercial solvent, is prepared by the reaction of \(\mathrm{Cl}_{2}(\mathrm{g})\) with a carbon compound. Determine \(\Delta H^{\circ}\) for the reaction $$ \mathrm{CS}_{2}(1)+3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{CCl}_{4}(1)+\mathrm{S}_{2} \mathrm{Cl}_{2}(1) $$ Use appropriate data from the following listing. $$\begin{aligned} \mathrm{CS}_{2}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{SO}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-1077 \mathrm{kJ} \end{aligned}$$ $$2 \mathrm{S}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{2} \mathrm{Cl}_{2}(1) \quad \Delta H^{\circ}=-58.2 \mathrm{kJ}$$ $$\mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{CCl}_{4}(1) \quad \Delta H^{\circ}=-135.4 \mathrm{kJ}$$ $$\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-296.8 \mathrm{kJ}$$ $$\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2} \mathrm{Cl}_{2}(1) \quad \Delta H^{\circ}=+97.3 \mathrm{kJ}$$ $$\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-393.5 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{CCl}_{4}(1)+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g})+\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) & \\ \Delta H^{\circ}=&-5.2 \mathrm{kJ} \end{aligned}$$

Can a chemical compound have a standard enthalpy of formation of zero? If so, how likely is this to occur? Explain.

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat \(=0.1291 \mathrm{Jg}^{-1} \mathrm{K}^{-1}\) ) with mass of \(10.5 \mathrm{g}\) is heated to \(78.3^{\circ} \mathrm{C}\) and immersed in \(50.0 \mathrm{g}\) of \(23.7^{\circ} \mathrm{C}\) water in a constant-pressure calorimeter. The final temperature of the water is \(31.0^{\circ} \mathrm{C}\). Is this a pure sample of gold?

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$
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