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 \(1 \mathrm{atm} :\) $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \Delta E^{\circ}=? $$ (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 given process at 298 K and 1 atm is approximately -284563.1 J/mol.

Step-by-step solution

Find the change in moles of gas

During the given reaction, the change in moles of gas produced is equal to the moles of hydrogen and oxygen gas produced minus the moles of liquid water reacted. We can write this as: $$ \Delta n = \text{moles of }H_2 + \frac{1}{2}\text{ moles of }O_2 - \text{moles of }H_2O $$ Since 1 mole of liquid water produces 1 mole of hydrogen gas and \(\frac{1}{2}\) moles of oxygen gas, the change in moles is: $$ \Delta n = 1 + \frac{1}{2} - 1 = \frac{1}{2} $$

Derive an expression for work in terms of \(n, R\), and \(T\)

For an ideal gas, the work done on the system during a reaction can be expressed as: $$ w = -P_{ext}\Delta V $$ Since we are given that the reaction occurs at constant temperature and pressure, we can apply the ideal gas equation which states that: $$ PV = nRT $$ We can rearrange this equation to solve for \(\Delta V\): $$ \Delta V =\frac{nRT}{P} $$ Next, we can express \(\Delta V\) in terms of the change in moles of gas, i.e. \(\Delta n\): $$ \Delta V = \Delta n\frac{RT}{P} $$ Now we can substitute this expression for \(\Delta V\) back into the work equation: $$ w = -P_{ext}\left(\Delta n\frac{RT}{P}\right) $$ Since the pressure is constant, we can cancel out the \(P_{ext}\) terms: $$ w = -\Delta nRT $$

Calculate the work done on the system

Substitute the values of \(\Delta n, R, T\) into the expression derived in Step 2: $$ w = -\left(\frac{1}{2}\right)(8.314\,\mathrm{J/mol\,K})(298\,\mathrm{K}) $$ Solving for the work done, we get: $$ w \approx -1236.9\,\mathrm{J/mol} $$

Convert the given enthalpy change in kJ/mol to J/mol

We are given the standard enthalpy change of formation for water as -285.8 kJ/mol. To maintain consistency in units, we need to convert this value to J/mol: $$ \Delta H = -285.8\,\mathrm{kJ/mol} \times 1000\,\mathrm{J/kJ} = -285800\,\mathrm{J/mol} $$

Calculate the change in internal energy

We can now use the enthalpy change, \(\Delta H\), and the work done, \(w\), to find the change in internal energy, \(\Delta E\), using the following formula: $$ \Delta E = \Delta H - w $$ Substitute the values of \(\Delta H\) and \(w\) into this equation: $$ \Delta E = -285800\,\mathrm{J/mol} - {-1236.9\,\mathrm{J/mol}} $$ Solving for the change in internal energy, we get: $$ \Delta E \approx -284563.1\,\mathrm{J/mol} $$ The change in internal energy for the given process at 298 K and 1 atm is approximately -284563.1 J/mol.

Key concepts

Enthalpy of Formation

Enthalpy of formation refers to the heat change that occurs when one mole of a compound is formed from its constituent elements in their standard states at a constant pressure.

This concept is crucial for understanding how energy is absorbed or released during chemical reactions. For water, \(\mathrm{H}_{2} \mathrm{O}(l)\), the standard enthalpy of formation is \(-285.8 \mathrm{kJ/mol}\).

An essential point to remember is that exothermic reactions, which release heat, have a negative enthalpy of formation value.

When calculating the change in internal energy (\(\Delta E\)) in a reaction, we can use the known value of enthalpy of formation along with work done on the system. This connection highlights the practical importance of enthalpy in determining energy dynamics in reactions.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates pressure, volume, temperature, and the amount of gas. The formula is given by:\(PV = nRT\), where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume,
• \(n\) is the number of moles of the gas,
• \(R\) is the ideal gas constant (8.314 J/mol K),
• \(T\) is the temperature in Kelvins.

This equation is especially useful for predicting the behavior of gases under various conditions. In the given exercise, the ideal gas law helps to calculate the volume change, which then allows us to find the work done during a reaction.

Remember that while the ideal gas law assumes perfect behavior, real gases may deviate slightly due to intermolecular forces and volume of gas particles, especially at high pressures or low temperatures.

Thermodynamic Work

Thermodynamic work refers to energy transferred when a system undergoes changes, such as expansion or compression. For ideal gases, work can be defined as:\(w = -P_{ext} \Delta V\), where \(\Delta V\) is the change in volume.

In the context of the ideal gas law:\(\Delta V = \Delta n \frac{RT}{P}\), where \(\Delta n\) is the change in moles.
This becomes significant in chemical reactions where gases expand or contract, altering internal energy.

In the given exercise, calculating \(w = -\Delta nRT\) allows us to determine the energy involved in gas expansion, contributing to the change in internal energy \(\Delta E\).

Understanding thermodynamic work is important for mastering how energy transfers occur in reactions and processes.

Unit Conversion in Chemistry

Unit conversion is an essential skill in chemistry, ensuring that all measurements are expressed in consistent units. This allows for accurate calculations and comparison of data.

For example, when dealing with energy, it's often necessary to convert kilojoules \(\mathrm{kJ}\) to joules \(\mathrm{J}\) since standard units are usually in the International System of Units \(\mathrm{SI}\).

In the given task, the enthalpy of formation is provided in \(\mathrm{kJ/mol}\), and is converted to \(\mathrm{J/mol}\) by multiplying by 1000.
This consistency is crucial for using the formula \(\Delta E = \Delta H - w\) accurately, as it ensures that all terms are calculated with the same units.

Mastering unit conversions will aid in solving a wide range of problems in chemistry and related sciences.