Question

The standard enthalpy of formation of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(298 \mathrm{K}\) is \(-285.8 \mathrm{kJ} / \mathrm{mol} .\) Calculate the change in internal energy for the following process at \(298 \mathrm{K}\) and 1 atm: $$\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \quad \Delta E^{\circ}=?$$ (Hint: Using the ideal gas equation, derive an expression for work in terms of \(n, R,\) and \(T .\) )

Short answer

The change in internal energy for the process \(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g)\) at \(298 \mathrm{K}\) and 1 atm is \(\Delta E^{\circ} = -285.8\, kJ/mol\).

Step-by-step solution

Derivation of Work Expression

Let's assume that the initial volume is \(V_1\) and the final volume is \(V_2\). The expression for work when the gas expands or contracts at constant temperature can be derived as follows: \(W = -\int_{V_1}^{V_2} P_{ext} dV\) Since pressure is constant at 1 atm, we can take it out of the integral, thus: \(W = -P_{ext}\int_{V_1}^{V_2} dV\) \(W = -P_{ext}(V_2 - V_1) = -P_{ext} \Delta V\) Now using the ideal gas equation, we can write the change in volume as: \(\Delta V = \frac{nR\Delta T}{P}\) Since the temperature is constant, \(\Delta T = 0\), which implies \(\Delta V = 0\) and hence the work done is also 0. 3. Calculate the change in internal energy using the given values. Now, we can use the relationship between enthalpy change, internal energy change, and work to solve for the internal energy change.

Calculate Internal Energy Change

Given: \(\Delta H^{\circ} = -285.8\, kJ/mol\), Work: \(W = 0\) We can use the formula: \(\Delta H = \Delta E + P \Delta V\) Since the work is 0, we have: \(\Delta H^{\circ} = \Delta E\) So, the change in internal energy is: \(\Delta E^{\circ} = -285.8\, kJ/mol\) The change in internal energy for the process H2O(l) => H2(g) + 1/2 O2(g) at 298 K and 1 atm is -285.8 kJ/mol.

Key concepts

Enthalpy Change

Enthalpy change, often symbolized as \(\Delta H\), is a measure of the total heat content of a system. It represents the energy transferred between the system and its surroundings at constant pressure. In the context of chemical reactions, the standard enthalpy of formation (like the \(\Delta H^\circ\) defined for the formation of \(\text{H}_2\text{O}(l)\)) is the change in enthalpy when one mole of a substance is formed from its elements under standard conditions (298 K and 1 atm).

The key relationship connecting enthalpy change to the internal energy change (\(\Delta E\)) and the work done by a gas is expressed by the equation: \(\Delta H = \Delta E + P\Delta V\). This is particularly useful when you want to determine one of these values when the other two are known. In the given problem, the enthalpy change serves as a direct measure of the change in internal energy since the work done at constant volume is zero, simplifying the relation to \(\Delta H^\circ = \Delta E^\circ\).

Internal Energy Change

Internal energy change, denoted as \(\Delta E\), is the total change in energy of a system. It's the sum of potential and kinetic energies which include chemical, thermal, and other forms of energy contained within the system. Internal energy can change through heat transfer or work being done on or by the system.

According to the first law of thermodynamics, which balances the energy accounts for any process, the change in internal energy is given by the equation: \(\Delta E = Q - W\), where \(Q\) is the heat added to the system, and \(W\) is the work done by the system. In our problem, since there is no work done by the gas (the volume change \(\Delta V\) is zero at constant temperature), the internal energy change is equal to the enthalpy change at constant volume, leading to \(\Delta E^\circ = -285.8\, kJ/mol\).

Ideal Gas Equation

The ideal gas equation, \(PV = nRT\), is a cornerstone in the study of thermodynamics and physical chemistry. It connects the pressure \(P\), volume \(V\), number of moles \(n\), universal gas constant \(R\), and temperature \(T\) for an ideal gas, which is a hypothetical gas that perfectly follows the kinetic-molecular theory.

In the equation, \(R\) is the ideal gas constant (8.314 J/(mol·K)) and the product \(nRT\) represents the energy associated with the motion of the moles of gas particles. The ideal gas equation allows us to derive expressions for work done on or by a gas when the system undergoes a change in terms of these variables. It is important to note that real gases only approximate the behavior predicted by the ideal gas law under certain conditions, but it is a useful model for many situations, including the calculation in our exercise.

Work Done by Gas

Work done by a gas during a process involving volume change is an essential concept in thermodynamics. It can be positive or negative, based on whether the gas expands or is compressed. Work (\(W\)) is calculated by the integral \(W = -\int_{V_1}^{V_2} P_{ext} dV\), where \(P_{ext}\) is the external pressure and \(V_1\) and \(V_2\) are the initial and final volumes, respectively.

However, in scenarios involving an ideal gas under conditions where temperature remains constant, the equation simplifies. Since \(\Delta T = 0\), no work is done (\(W = 0\)), as there is no volume change. In the provided exercise, this gives us a clear relation between the standard enthalpy of formation and internal energy change, since all the energy change can be attributed to internal energy, with no work contributing to the energy transfer.