Question

A. One hundred milliliters of \(0.200 \mathrm{M}\) ATP is mixed with an ATPase in a Dewar at \(298 \mathrm{~K}, 1\) atm, \(\mathrm{pH} 7.0, \mathrm{pMg} 3.0\), and \(0.25 \mathrm{M}\) ionic strength. The temperature of the solution increases \(1.48 \mathrm{~K}\). What is \(\Delta H^{*}\) for the hydrolysis of ATP to adenosine 5'-diphosphate (ADP) and phosphate? Assume the heat capacity of the system is \(418 \mathrm{~J} / \mathrm{K}\). B. The hydrolysis reaction can be written as $$ \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{ADP}+\mathrm{P}_{\mathrm{i}} $$ Under the same conditions, the hydrolysis of ADP, $$ \mathrm{ADP}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{AMP}+\mathrm{P}_{\mathrm{i}} $$ has a heat of reaction, \(\Delta H^{\circ}\), of \(-28.9 \mathrm{~kJ} / \mathrm{mol}\). Under the same conditions, calculate \(\Delta H^{\circ}\) for the adenylate kinase reaction: $$ 2 \mathrm{ADP} \rightleftharpoons \mathrm{AMP}+\mathrm{ATP} $$

Short answer

A: \( \Delta H^{*} = 30.93 \ \text{kJ/mol} \). B: \( \Delta H^{\circ} = -59.83 \ \text{kJ/mol} \).

Step-by-step solution

Understand the Problem

For Part A, we need to find the enthalpy change of the reaction, \( \Delta H^{*} \), when ATP is hydrolyzed into ADP and phosphate. The given temperature rise and heat capacity will be used to determine it. For Part B, we aim to calculate \( \Delta H^{\circ} \) for an adenylate kinase reaction using given enthalpies of related reactions.

Calculate Enthalpy Change Using Temperature Change (Part A)

The equation relating heat change, \( q \), to enthalpy change is \( q = \Delta H = C_p \cdot \Delta T \), where \( C_p \) is the heat capacity and \( \Delta T \) is the temperature change. Here, \( C_p = 418 \ \text{J/K} \) and \( \Delta T = 1.48 \ \text{K} \). Thus, \( q = 418 \times 1.48 \ \text{J} = 618.64 \ \text{J} \). Since the solution is 100 ml of 0.2 M ATP, moles of ATP are \( (100 \times 10^{-3}) \times 0.2 = 0.02 \) mol.

Calculate \( \Delta H^{*} \) for Hydrolysis Reaction (Part A)

Now calculate \( \Delta H^{*} \) per mole of ATP: \( \Delta H^{*} = \frac{618.64 \ \text{J}}{0.02 \ \text{mol}} = 30932 \ \text{J/mol} \) or approximately \( 30.93 \ \text{kJ/mol} \).

Set Up Reaction Enthalpy Relationships (Part B)

In Part B, the heat of reactions are algebraically related. We know: 1. \( \text{ATP} + \text{H}_2\text{O} \rightleftharpoons \text{ADP} + \text{P}_i \) with \( \Delta H^{\circ} = 30.93 \ \text{kJ/mol} \) calculated in Part A.2. \( \text{ADP} + \text{H}_2\text{O} \rightleftharpoons \text{AMP} + \text{P}_i \) with \( \Delta H^{\circ} = -28.9 \ \text{kJ/mol} \).3. The target: \( 2 \text{ADP} \rightleftharpoons \text{AMP} + \text{ATP} \).

Apply Hess's Law (Part B)

Using Hess's Law, the reactions can be manipulated to solve for the needed enthalpy. Combine the first reaction in reverse, the second as is, and cancel out terms:- Reverse Reaction 1: \( \text{ADP} + \text{P}_i \rightleftharpoons \text{ATP} + \text{H}_2\text{O} \) with \( \Delta H^{\circ} = -30.93 \ \text{kJ/mol} \).- Add Reaction 2: \( \text{ADP} + \text{H}_2\text{O} \rightleftharpoons \text{AMP} + \text{P}_i \) with \( \Delta H^{\circ} = -28.9 \ \text{kJ/mol} \).- \( -30.93 \ \text{kJ/mol} + (-28.9 \ \text{kJ/mol}) = \Delta H^{\circ}(\text{adenylate kinase}) = -59.83 \ \text{kJ/mol}.\)

Key concepts

Enthalpy Change Calculation

The enthalpy change, represented as \( \Delta H \), is a measurement of heat transfer during a chemical reaction at constant pressure. In biochemical systems, understanding \( \Delta H \) is crucial for energetics of reactions. In our scenario, we compute the enthalpy change for the hydrolysis of ATP to ADP and phosphate. The process involves a rise in temperature indicating an exothermic reaction.
To find \( \Delta H \), use the formula \( q = \Delta H = C_p \cdot \Delta T \), where \( C_p \) is the heat capacity and \( \Delta T \) is the temperature change observed. Here, as given, \( C_p = 418 \ \text{J/K} \) and \( \Delta T = 1.48 \ \text{K} \). Consequently, calculate the heat absorbed or released as \( q = 618.64 \ \text{J} \).
Since the solution has 0.02 moles of ATP, the enthalpy change per mole is \( \Delta H^{*} = \frac{618.64 \ \text{J}}{0.02 \ \text{mol}} = 30932 \ \text{J/mol} \), which simplifies to about \( 30.93 \ \text{kJ/mol} \). This value signifies the energy involved per mole of ATP hydrolyzed.

Adenylate Kinase Reaction

Adenylate Kinase plays a significant role in cellular energy homeostasis, catalyzing the interconversion of adenine nucleotides. In the given exercise, we address the specific reaction \( 2 \ \text{ADP} \rightleftharpoons \text{AMP} + \text{ATP} \). Understanding this reaction helps illustrate resource recycling within cells.
The challenge is to find the enthalpy change, \( \Delta H^{\circ} \), for this transformation. Given the enthalpy changes for related reactions, we use principles that simplify complex biochemical pathways into manageable steps. Knowing \( \Delta H^{\circ} \) helps predict the spontaneity and feasibility of biochemical pathways in vivo.

Hess's Law in Biochemistry

Hess's Law is instrumental in thermodynamics, stating that the overall enthalpy change of a chemical reaction is independent of the pathway taken. This law is incredibly useful in biochemistry where direct measurement of some reactions may be challenging.
In our exercise, Hess's Law helps calculate the enthalpy change for adenylate kinase reaction. To achieve this, reverse Reaction 1 (ATP hydrolysis), rechannelling \( \,\text{ADP} + \text{P}_i \rightleftharpoons \,\text{ATP} + \text{H}_2\text{O} \) with \( \Delta H^{\circ} = -30.93 \ \text{kJ/mol} \), and add it to Reaction 2, leaving \( \,\text{ADP} + \text{H}_2\text{O} \rightleftharpoons \,\text{AMP} + \text{P}_i \) at \( \Delta H^{\circ} = -28.9 \ \text{kJ/mol} \).
When summed, the overall \( \Delta H^{\circ} = -59.83 \ \text{kJ/mol} \) for the targeted adenylate kinase reaction, demonstrating the efficiency of Hess's Law in biochemical thermodynamics.