Question

Used in welding metals, the reaction of acetylene with oxygen has $$\begin{aligned}\Delta H^{\circ}=&-1256.2 \mathrm{~kJ}: \\ \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+5 / 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \\ \Delta H^{\circ}=-1256.2 \mathrm{~kJ}\end{aligned}$$ How much \(P V\) work is done in kilojoules and what is the value of \(\Delta E\) in kilojoules for the reaction of \(6.50 \mathrm{~g}\) of acetylene at atmospheric pressure if the volume change is \(-2.80 \mathrm{~L}\) ?

Short answer

Work done: 0.284 kJ, \(\Delta E = -313.43\) kJ.

Step-by-step solution

Calculate Moles of Acetylene

First, determine the number of moles of acetylene (\(\mathrm{C}_2\mathrm{H}_2\)) using its molar mass. The molar mass of \(\mathrm{C}_2\mathrm{H}_2\) is approximately 26.04 g/mol. Given 6.50 g of acetylene, the moles \(n\) is calculated as follows:\[n = \frac{\text{mass}}{\text{molar mass}} = \frac{6.50}{26.04} \approx 0.2497 \text{ moles}\]

Calculate Work Done (\(P\Delta V\))

The work done by the system at constant pressure can be calculated using the formula: \(W = -P\Delta V\).Given that the pressure \(P\) is 1 atm (converted to 101.325 J/L for work calculation) and the volume change \(\Delta V\) is \(-2.80\) L, the work done is:\[W = -101.325 \times (-2.80) = 283.71 \text{ J} \quad \text{or} \quad 0.2837 \text{ kJ}\]

Calculate Enthalpy Change for the Given Reaction

The given reaction's enthalpy change \(\Delta H^{\circ}\) for one mole of acetylene is \(-1256.2\) kJ. For the reaction using \(0.2497\) moles of acetylene, the enthalpy change is calculated by multiplying:\[\Delta H = -1256.2 \times 0.2497 \approx -313.71 \text{ kJ}\]

Calculate Change in Internal Energy (\(\Delta E\))

The change in internal energy \(\Delta E\) can be calculated using the relationship:\(\Delta E = \Delta H + W\).Substitute the values calculated:\[\Delta E = -313.71 + 0.2837 \approx -313.43 \text{ kJ}\]

Conclusion

The PV work done is approximately 0.284 kJ, and the change in internal energy \(\Delta E\) for the reaction is approximately \(-313.43\) kJ.

Key concepts

Enthalpy Change

In the study of thermochemistry, enthalpy change (\( \Delta H \)) is a fundamental concept that helps us understand how heat energy is involved in chemical reactions. Enthalpy change describes the total energy change within a system during a reaction and is typically measured in kilojoules (kJ).

When a reaction releases energy, it is known as an exothermic process, and \( \Delta H \) is negative, indicating that heat exits the system. Conversely, endothermic reactions absorb energy, resulting in a positive \( \Delta H \) value. For the reaction of acetylene with oxygen, the given \( \Delta H^{\circ} \) is \(-1256.2\text{ kJ} \) for one mole of acetylene, signifying a highly exothermic reaction. This is why it is effective in welding, providing the needed heat to meld metals.

• Enthalpy can be affected by the state of reactants and products, such as whether water is in liquid or vapor form.
• Standard conditions (indicated by the degree symbol in \( \Delta H^{\circ} \)) are usually 298 K and 1 atm.

The calculation in the exercise involves adjusting this value according to the moles of reactants involved to find the specific energy change during the actual reaction scenario discussed.

Internal Energy

Internal energy (\( \Delta E \)) is another vital aspect of thermochemistry, describing the total energy contained within a system. It takes into account both the thermal energy (heat changes) and the work done during the reaction process.

This concept is linked to the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system plus the work done on the system. Mathematically, it is expressed as:\[ \Delta E = \Delta H + W \]where \( W \) is the work done on or by the system. In the context of the acetylene reaction, while the reaction itself releases a large amount of heat (\(\Delta H \) being negative), the actual energy change \( \Delta E \) also factors in any work associated with volume change (discussed in the "PV Work" section).

• Internal energy is a state function, meaning it is determined by the state of the system, not how it got there.
• Both heat and work are paths dependent; however, their sum isn't.

Understanding internal energy changes helps clarify how much of the reaction energy is accessible for different purposes, such as mechanical work or heat production.

PV Work

In chemical thermodynamics, "PV work" refers to the work done by a system when it expands or contracts against an external pressure during a reaction. The formula to calculate this work is:
\[ W = -P\Delta V \]
where:

• \( P \) is the external pressure.
• \( \Delta V \) is the change in volume (final - initial).

The negative sign in the formula indicates that when the system expands (\( \Delta V > 0 \)), it does work on the surroundings, and thus, energy is lost. Conversely, when the system contracts (\( \Delta V < 0 \)), work is done on the system, and energy is gained.

In the welding reaction using acetylene, with a volume change of \(-2.80 \text{ L}\), we calculated the work done as approximately \(0.2837 \text{ kJ}\) given atmospheric pressure (1 atm converted into appropriate units for energy calculation).
This small work value shows that most of the energy change in this reaction is dominated by heat changes rather than mechanical work, which is characteristic of many gas reactions. Understanding PV work is crucial for accounting for how pressure and volume changes influence a system's total energy exchange.