Assume that a particular reaction evolves 244 kJ of heat and that 35 kJ of PV work is gained by the system. What are the values of ∆E and ∆H for the system? For the surroundings?

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Identify the given values: the heat evolved by the reaction is 244 kJ, and the PV work done on the system is 35 kJ.

Recall the first law of thermodynamics, which states that the change in internal energy (\( \Delta E \)) of a system is the sum of the heat exchanged (q) and the work done (w) on the system: \( \Delta E = q + w \).

Since the reaction evolves heat, the heat exchanged (q) is -244 kJ (exothermic process). The work done on the system (w) is +35 kJ.

Substitute the values into the equation \( \Delta E = q + w \) to find the change in internal energy for the system.

To find the change in enthalpy (\( \Delta H \)), use the relation \( \Delta H = \Delta E + P\Delta V \). Since \( P\Delta V \) is the work done, \( \Delta H = q \) for processes at constant pressure, so \( \Delta H = -244 \text{ kJ} \).

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First Law of Thermodynamics

The First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed from one form to another. In a closed system, the change in internal energy (∆E) is equal to the heat added to the system minus the work done by the system. This principle is fundamental for understanding how energy transfers occur during chemical reactions.

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In thermodynamics, work and heat are two primary ways energy can be transferred between a system and its surroundings. Work (W) is the energy transferred when a force is applied over a distance, while heat (q) is the energy transferred due to a temperature difference. Understanding how to account for these forms of energy transfer is crucial for calculating changes in internal energy and enthalpy in chemical reactions.

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