Question

For the hydrolysis of ATP at \(25^{\circ} \mathrm{C}(298 \mathrm{K})\) and \(\mathrm{pH} 7, \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{ADP}+\mathrm{P}_{1}+\mathrm{H}^{+},\) the standard free energy of hydrolysis \(\left(\Delta G^{\circ}\right)\) is \(-30.5 \mathrm{kJ} \mathrm{mol}^{-1}\left(-7.3 \mathrm{kcal} \mathrm{mol}^{-1}\right),\) and the standard enthalpy change \(\left(\Delta H^{\circ \prime}\right)\) is \(-20.1 \mathrm{kJ} \mathrm{mol}^{-1}\left(-4.8 \mathrm{kcal} \mathrm{mol}^{-1}\right)\) Calculate the standard entropy change \(\left(\Delta S^{\circ}\right)\) for the reaction, in both joules and calories. Why is the positive sign of the answer to be expected in view of the nature of the reaction? Hint: You may want to review some material from Chapter 1.

Short answer

\(\Delta S^{\circ} \approx 34.9 \, \mathrm{J} \, \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \approx 8.34 \, \mathrm{cal} \, \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\). The positive sign is expected due to the increased disorder.

Step-by-step solution

Understand the Gibbs Free Energy Equation

To find the standard entropy change \(\Delta S^{\circ}\), we use the Gibbs Free Energy equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\]

Rearrange the Equation

We need to solve for \(\Delta S^{\circ}\). Rearrange the equation to: \[\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T}\]

Insert Known Values Into the Equation

Given values: \(\Delta G^{\circ} = -30.5\, \mathrm{kJ} \, mol^{-1}\) and \(\Delta H^{\circ} = -20.1\, \mathrm{kJ} \, mol^{-1}\), and \(T = 298 \, \mathrm{K}\). Substitute these into the equation: \[\Delta S^{\circ} = \frac{-20.1 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1} + 30.5 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}}{298 \, \mathrm{K}}\]

Calculate the Standard Entropy Change in Joules

Perform the calculation: \[\Delta S^{\circ} = \frac{10.4 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}}{298 \, \mathrm{K}} \approx 0.0349 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\] Convert \(\mathrm{kJ} \, \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\) to \(\mathrm{J} \, \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\): \[\Delta S^{\circ} \approx 34.9 \, \mathrm{J} \, \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\]

Convert the Entropy Change to Calories

To convert from joules to calories, use the conversion factor \(1 \, \mathrm{cal} = 4.184 \, \mathrm{J}\): \[\Delta S^{\circ} = \frac{34.9 \, \mathrm{J}}{4.184} \approx 8.34 \, \mathrm{cal} \, \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\]

Interpret the Sign of \(\Delta S^{\circ}\)

The positive sign of \(\Delta S^{\circ}\) indicates that entropy increases during the reaction. This is expected because the hydrolysis of ATP results in the formation of more molecules (ADP, \(P_1\), and \(H^+\)), thus increasing disorder.

Key concepts

Gibbs Free Energy equation

Understanding the Gibbs Free Energy equation is crucial in thermodynamics, especially in biochemistry. The equation is: \[\begin{equation} \Delta G^{\text{\circ}} = \Delta H^{\text{\circ}} - T \Delta S^{\text{\circ}} \end{equation}\]This equation relates Gibbs free energy (\[\begin{equation} \Delta G^{\text{\circ}} \end{equation}\]), enthalpy (\[\begin{equation} \Delta H^{\text{\circ}} \end{equation}\]), temperature (\[\begin{equation} T \end{equation}\]), and entropy (\[\begin{equation} \Delta S^{\text{\circ}} \end{equation}\]). Essentially, it tells us whether a reaction will happen spontaneously. A negative \[\begin{equation} \Delta G^{\text{\circ}} \end{equation}\] means the reaction is spontaneous. By rearranging this equation, you can find other values like entropy.

Entropy change

Entropy is a measure of disorder or randomness in a system. For the hydrolysis of ATP, calculating the entropy change (\[\begin{equation} \Delta S^{\text{\circ}} \end{equation}\] can show how the disorder changes during the reaction. You can use the Gibbs Free Energy equation and solve for \[\begin{equation} \Delta S^{\text{\circ}} \end{equation}\]: \[\begin{equation} \Delta S^{\text{\circ}} = \frac {\Delta H^{\text{\circ}} - \Delta G^{\text{\circ}}} { T } \end{equation}\] Plugging in known values \[\begin{equation} \Delta G^{\text{\circ}} = -30.5 \ \text{kJ mol}^{-1} \end{equation}\], \[\begin{equation} \Delta H^{\text{\circ}} = -20.1 \ \text{kJ mol}^{-1} \end{equation}\], and \[\begin{equation} T = 298 \text{K} \end{equation}\], the calculation gives: \[\begin{equation} \Delta S^{\text{\circ}} = \frac{10.4 \text{kJ mol}^{-1}}{298 \text{K}} \approx 0.0349 \text{kJ mol}^{-1 K}^{-1} \end{equation}\]. Convert it to \[\begin{equation} \text{J mol}^{-1 K}^{-1} = 34.9 \text{J mol}^{-1 K}^{-1} \end{equation}\].

Standard free energy of hydrolysis

The standard free energy of hydrolysis (\[\begin{equation} \Delta G^{\text{\circ}} \end{equation}\]) is the change in Gibbs free energy when a compound is hydrolyzed under standard conditions. For ATP hydrolysis, this value is significantly negative (\[\begin{equation} -30.5 \ \text{kJ mol}^{-1} \end{equation}\]), indicating the reaction proceeds spontaneously. This makes ATP a crucial energy currency in cells, as its breakdown releases energy that can power various cellular processes.

Standard enthalpy change

Enthalpy (\[\begin{equation} \Delta H^{\text{\circ}} \end{equation}\]) measures the heat change at constant pressure. For ATP hydrolysis, the standard enthalpy change is-\[\begin{equation} 20.1 \ \text{kJ mol}^{-1} \end{equation}\]. This negative value means the reaction releases heat and is exothermic. The enthalpy change helps determine the total energy change, which, combined with entropy, influences the Gibbs free energy.

Thermodynamics in biochemistry

Thermodynamics in biochemistry deals with energy transfer and transformations within living organisms. The principles of thermodynamics, such as Gibbs free energy, enthalpy, and entropy, are vital for understanding biochemical reactions. For instance: * **Gibbs free energy**: Tells whether a reaction is spontaneous* **Enthalpy**: Measures heat changes during reactions* **Entropy**: Indicates the randomness or disorder change in a system Understanding these concepts is essential for studying cellular processes, energy metabolism, and enzyme mechanisms.