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Chapter 2: Problem 2

The alcohol dehydrogenase reaction, which removes ethanol from your blood, proceeds according to the following reaction: \(\mathrm{NAD}^{+}+\)Ethanol \(\rightleftarrows \mathrm{NADH}+\) Acetaldehyde Under standard conditions ( \(298 \mathrm{~K}, 1 \mathrm{~atm}, \mathrm{pH} 7.0, \mathrm{pMg} 3\), and an ionic strength of \(0.25 \mathrm{M}\) ), the standard enthalpies and free energies of formation of the reactants are as follows: \begin{tabular}{lrr} & \(H^{\circ}(\mathrm{kJ} / \mathrm{mol})\) & \(G^{\circ}(\mathrm{kJ} / \mathrm{mol})\) \\ \hline NAD \(^{+}\) & \(-10.3\) & \(1059.1\) \\ NADH & \(-41.4\) & \(1120.1\) \\ Ethanol & \(-290.8\) & \(63.0\) \\ Acetaldehyde & \(-213.6\) & \(24.1\) \\ \hline \end{tabular} A. Calculate \(\Delta G^{\circ}, \Delta H^{\circ}\), and \(\Delta S^{\circ}\) for the alcohol dehydrogenase reaction under standard conditions. B. Under standard conditions, what is the equilibrium constant for this reaction? Will the equilibrium constant increase or decrease as the temperature is increased?

Short Answer

Expert verified
\( \Delta G^{\circ} = 118.0 \, \text{kJ/mol}, \Delta H^{\circ} = 46.1 \, \text{kJ/mol}, \Delta S^{\circ} = -241.6 \, \text{J/mol K}, K_{eq} \approx 2.92 \times 10^{-21} \). The equilibrium constant increases with temperature.

Step by step solution

01

Calculate \( \Delta G^{\circ} \) For Reaction

The Gibbs free energy change for the reaction, \( \Delta G^{\circ} \), can be calculated using the formula:\[ \Delta G^{\circ} = \sum G^{\circ}_{\text{products}} - \sum G^{\circ}_{\text{reactants}} \]For the reaction \( \mathrm{NAD}^{+} + \text{Ethanol} \rightarrow \mathrm{NADH} + \text{Acetaldehyde} \), plug in the given values:\[ \Delta G^{\circ} = (1120.1 + 24.1) - (1059.1 + 63.0) = 118.0 \, \text{kJ/mol} \]
02

Calculate \( \Delta H^{\circ} \) For Reaction

The enthalpy change for the reaction, \( \Delta H^{\circ} \), can be calculated using the formula:\[ \Delta H^{\circ} = \sum H^{\circ}_{\text{products}} - \sum H^{\circ}_{\text{reactants}} \]Plug in the values for the reaction:\[ \Delta H^{\circ} = (-41.4 - 213.6) - (-10.3 - 290.8) = 46.1 \, \text{kJ/mol} \]
03

Calculate \( \Delta S^{\circ} \) Using \( \Delta G^{\circ} \) and \( \Delta H^{\circ} \)

Use the relation between Gibbs free energy, enthalpy, and entropy:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]Rearranging the formula gives:\[ \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} \]Using \( T = 298 \, K \), substitute the values:\[ \Delta S^{\circ} = \frac{46.1 - 118.0}{298} = -0.2416 \, \text{kJ/mol K} = -241.6 \, \text{J/mol K} \]
04

Calculate the Equilibrium Constant \( K_{eq} \)

The equilibrium constant \( K_{eq} \) is found using the relation:\[ \Delta G^{\circ} = -RT \ln K_{eq} \]Rearranging gives:\[ \ln K_{eq} = -\frac{\Delta G^{\circ}}{RT} \]With \( R = 8.314 \, \text{J/mol K} \) and \( T = 298 \, K \), calculate:\[ \ln K_{eq} = -\frac{118000}{8.314 \times 298} \approx -47.49 \]\[ K_{eq} = e^{-47.49} \approx 2.92 \times 10^{-21} \]
05

Analyze Effect of Temperature on Equilibrium Constant

The reaction is endothermic \((\Delta H^{\circ} > 0)\), so according to Le Chatelier's principle, an increase in temperature will shift the equilibrium to favor products. This means \( K_{eq} \) will increase with temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs free energy, denoted as \( \Delta G^{\circ} \), helps us determine the spontaneity of a chemical reaction. It combines both enthalpy (heat content) and entropy (disorder) changes in a system to predict whether a reaction will occur under given conditions. Mathematically, it is expressed as \( \Delta G^{\circ} = \sum G^{\circ}_{\text{products}} - \sum G^{\circ}_{\text{reactants}} \). If the value of \( \Delta G^{\circ} \) is negative, the reaction proceeds spontaneously, meaning it releases energy and occurs naturally. However, a positive \( \Delta G^{\circ} \) indicates that the reaction is non-spontaneous and requires energy input to proceed.
In biological systems, understanding Gibbs free energy is crucial because it helps predict the direction and extent of metabolic reactions, aiding in understanding how cells utilize energy.
Enthalpy Change
Enthalpy change, expressed as \( \Delta H^{\circ} \), represents the heat change during a reaction at constant pressure. It is an important thermodynamic parameter that can indicate whether a reaction absorbs or releases heat. You calculate it using the formula: \( \Delta H^{\circ} = \sum H^{\circ}_{\text{products}} - \sum H^{\circ}_{\text{reactants}} \).
  • A positive \( \Delta H^{\circ} \) implies an endothermic reaction, which absorbs heat from the surroundings.
  • A negative \( \Delta H^{\circ} \) indicates an exothermic reaction, which releases heat.
\( \Delta H^{\circ} \) affects the behavior of reactions, and in biochemistry, it helps in understanding how much energy is needed or released during metabolism and how organisms maintain their internal energy balance.
Entropy Change
Entropy change \( (\Delta S^{\circ}) \) measures the change in disorder or randomness in a system during a reaction. It is calculated using the relationship between Gibbs free energy, enthalpy, and temperature: \[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]Rearranging gives: \[ \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} \]
A positive \( \Delta S^{\circ} \) suggests increasing disorder, which typically favors spontaneity. Conversely, a negative \( \Delta S^{\circ} \) implies decreasing disorder, which may not favor the reaction unless energy is provided. In biological systems, entropy reflects how systems move towards more disordered states spontaneously, impacting how cells manage energy and catalyze reactions efficiently.
Equilibrium Constant
The equilibrium constant \( K_{eq} \) quantifies the ratio of products to reactants at equilibrium, providing insight into the position of equilibrium in a chemical reaction. It is related to Gibbs free energy by the equation: \[ \Delta G^{\circ} = -RT \ln K_{eq} \]where R is the gas constant and T is the temperature in Kelvin.
  • If \( K_{eq} > 1 \), the reaction favors products at equilibrium.
  • If \( K_{eq} < 1 \), the reaction favors reactants.
\[ K_{eq} \] is temperature-dependent, and based on Le Chatelier's principle, understanding its variations with temperature variations helps predict how reactions will behave under different thermal conditions. In biology, \( K_{eq} \) can influence how metabolic pathways are regulated and how efficiently substrates are converted into necessary compounds for survival.

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Most popular questions from this chapter

What is the maximum amount of work that can be obtained from hydrolyzing 1 mole of ATP to ADP and phosphate at \(298 \mathrm{~K}\) ? Assume that the concentrations of all reactants are \(0.01 \mathrm{M}\) and \(\Delta G^{\circ}\) is \(-32.5 \mathrm{~kJ} / \mathrm{mol}\). If the conversion of free energy to mechanical work is \(100 \%\) efficient, how many moles of ATP would have to be hydrolyzed to lift a \(100 \mathrm{~kg}\) weight 1 meter high?

A. It has been proposed that the reason ice skating works so well is that the pressure from the blades of the skates melts the ice. Consider this proposal from the viewpoint of phase equilibria, The phase change in question is $$ \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(/) $$ Assume that \(\Delta H\) for this process is independent of temperature and pressure and is cqual to 80 cal/g. The change in volume, \(\Delta V\), is about \(-9.1 \times\) \(10^{-5} \mathrm{~L} / \mathrm{g}\). The pressure exerted is the force per unit area. For a 180 pound person and an area for the skate blades of about 6 square inches, the pressure is \(30 \mathrm{lb} / \mathrm{sq}\). in. or about 2 atmospheres. With this information, calculate the decrease in the melting temperature of ice caused by the skate blades. (Note that 1 cal \(=0.04129\) L.atm.) Is this a good explanation for why ice skating works? B. A more efficient way of melting ice is to add an inert compound such as urea. (We will avoid salt to save our cars.) The extent to which the freexing point is lowered can be calculated by noting that the molar free energy of water must be the same in the solid ice and the urea solution. The molar free energy of water in the urea solution can be approximated as \(G_{\text {liquid }}^{\circ}+R T \ln X_{\text {water }}\) where \(X_{\text {water }}\) is the mole fraction of water in the solution. The molar free energy of the solid can be written as \(G_{\text {solid }}^{\circ}\). Derive an expression for the change in the melting temperature of ice by equating the free energies in the two phases, differentiating the resulting equation with respect to temperature, integrating from a mole fraction of 1 (pure solvent) to the mole fraction of the solution, and noting that \(\ln X_{\text {wamer }}=\ln \left(1-X_{\text {urea }}\right)\) \(=-X_{\text {urea }}\) (This relationship is the series expansion of the logarithm for small values of \(X_{\text {unea }}\). Since the concentration of water is about \(55 \mathrm{M}\), this is a good approximation.) With the relationship derived, estimate the de- crease in the melting temperature of ice for an \(1 \mathrm{M}\) urea solution. The heat of fusion of water is \(1440 \mathrm{cal} / \mathrm{mol}\).

The equilibrium constant under standard conditions \((1 \mathrm{~atm}, 298 \mathrm{~K}, \mathrm{pH} 7,0)\) for the reaction catalyzed by fumarase, Fumarate \(+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{L}\)-malate is \(4.00\). At \(310 \mathrm{~K}\), the equilibrium constant is \(8.00\). A. What is the standard free energy change, \(\Delta G^{\circ}\), for this reaction? B. What is the standard enthalpy change, \(\Delta H^{\circ}\), for this reaction? Assume the standard enthalpy change is independent of temperature. C. What is the standard entropy change, \(\Delta 5^{\circ}\), at \(298 \mathrm{~K}\) for this reaction? D. If the concentration of both reactants is equal to \(0.01 \mathrm{M}\), what is the free energy change at \(298 \mathrm{~K}\) ? As the reaction proceeds to equilibrium, will more fumarate or L-malate form? E. What is the free energy change for this reaction when equilibrium is reached?

The following reaction is catalyzed by the enzyme creatine kinase: A. Under standard conditions \((1 \mathrm{~atm}, 298 \mathrm{~K}, \mathrm{pH} 7.0 . \mathrm{pMg} 3.0\) and an ionic strength of \(0.25 \mathrm{M}\) ) with the concentrations of all reactants equal to \(10 \mathrm{mM}\), the free energy change, \(\Delta G\), for this reaction is \(13.3 \mathrm{~kJ} / \mathrm{mol}\). What is the standard free energy change, \(\Delta G^{\circ}\), for this reaction? B. What is the equilibrium constant for this reaction? C. The standard enthalpies of formation for the reactants are as follows: \(\begin{array}{ll}\text { Creatine } & -540 \mathrm{~kJ} / \mathrm{mol} \\\ \text { Creatine phosphate } & -1510 \mathrm{~kJ} / \mathrm{mol} \\ \text { ATP } & -2982 \mathrm{~kJ} / \mathrm{mol} \\ \text { ADP } & -2000 \mathrm{~kJ} / \mathrm{mol}\end{array}\) What is the standard enthalpy change, \(\Delta H^{\circ}\), for this reaction? D. What is the standard entropy change, \(\Delta S^{\circ}\), for this reaction?
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