Question

The addition of \(\mathrm{H}_{2}\) to \(\mathrm{C}=\mathrm{C}\) double bonds is an important reaction used in the preparation of margarine from vegetable oils. If \(50.0 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(50.0 \mathrm{~mL}\) of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) are allowed to react at \(1.5\) atm, the product ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) has a volume of \(50.0 \mathrm{~mL}\). Calculate the amount of \(P V\) work done, and tell the direction of the energy flow. $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$$

Short answer

Work done: 7.6 J; energy flowed out of the system.

Step-by-step solution

Identify the Reaction and Conditions

We have a gas-phase addition reaction: \(\mathrm{C}_{2}\mathrm{H}_4(g) + \mathrm{H}_2(g) \rightarrow \mathrm{C}_{2}\mathrm{H}_6(g)\). Initial volumes of \(\mathrm{H}_2\) and \(\mathrm{C}_{2}\mathrm{H}_4\) are both \(50.0\, \mathrm{mL}\) at \(1.5\, \mathrm{atm}\). The product \(\mathrm{C}_{2}\mathrm{H}_6\) is \(50.0\, \mathrm{mL}\). We need to calculate the work done and determine the direction of energy flow.

Calculate Volume Change

The initial total volume of reactants is \(50.0\, \mathrm{mL} + 50.0\, \mathrm{mL} = 100.0\, \mathrm{mL}\). The final volume of the product is \(50.0\, \mathrm{mL}\). Thus, the change in volume \(\Delta V\) is \(50.0\, \mathrm{mL} - 100.0\, \mathrm{mL} = -50.0\, \mathrm{mL}\).

Convert Volume to Liters

Convert \(\Delta V\) from milliliters to liters: \(-50.0\, \mathrm{mL} = -0.0500\, \mathrm{L}\).

Calculate PV Work Done

The work done by a gas is given by the equation \(w = -P\Delta V\). Here \(P = 1.5\, \mathrm{atm}\) and \(\Delta V = -0.0500\, \mathrm{L}\). Substituting these values, we get:\[w = -1.5\, \mathrm{atm} \times (-0.0500\, \mathrm{L}) = 0.075\, \mathrm{L}\cdot\mathrm{atm}\].

Convert Work to Joules

To convert work from \(\mathrm{L}\cdot\mathrm{atm}\) to joules, use the conversion factor \(1\, \mathrm{L}\cdot\mathrm{atm} = 101.3\, \mathrm{J}\). Therefore:\[w = 0.075\, \mathrm{L}\cdot\mathrm{atm} \times 101.3\, \mathrm{J/L}\cdot\mathrm{atm} = 7.5975\, \mathrm{J}\].

Determine Direction of Energy Flow

Since the system did work on the surroundings (\(-P\Delta V\) was negative), energy flowed out of the system.

Key concepts

PV Work

In thermodynamics, PV work refers to the work done by or on a gas during a reaction where volume change occurs at constant pressure. This often happens in chemical reactions involving gases. In our reaction, ethylene and hydrogen react to form ethane—all in the gaseous state. The equation for calculating work (\(w \)) performed by a gas is: \[w = -P \Delta V\]where

• \(P\) is the pressure of the gas,
• \(\Delta V\) is the change in volume.

To calculate PV work during the reaction from ethylene (\(\mathrm{C}_2\mathrm{H}_4\)) and hydrogen (\(\mathrm{H}_2\)) to ethane (\(\mathrm{C}_2\mathrm{H}_6\)), we note that the initial total volume of gases was 100.0 mL, and the final volume of the resultant ethane was 50.0 mL. With a pressure of 1.5 atm, PV work can first be calculated in liter-atmospheres and then converted to joules for clarity. The negative sign in the equation reflects the direction of work energy flow, emphasizing whether energy is added to or removed from the system.

Volume Change in Chemical Reactions

Volume change is key in understanding reactions involving gases. It's expressed as \(\Delta V\). Initially, the combined volume of the reactants (\(\mathrm{C}_2\mathrm{H}_4\) and \(\mathrm{H}_2\)) was 100.0 mL. This reduced to 50.0 mL when converted to ethane (\(\mathrm{C}_2\mathrm{H}_6\)).Thus, the volume change is calculated as:\[\Delta V = V_{\text{final}} - V_{\text{initial}} = 50.0\, \text{mL} - 100.0\, \text{mL} = -50.0\, \text{mL}\]Since the result is negative, it shows a decrease in volume. Converting this to liters gives -0.0500 L. Volume reduction in a reaction often involves gases being absorbed or transformed into fewer gaseous molecules in the product. For reactions happening at constant pressure, these changes significantly impact the work done during the reaction, termed PV work.

Energy Flow Direction

The determination of energy flow direction is crucial in chemical processes. It indicates whether energy leaves or enters the system. For the reaction of ethylene and hydrogen gases to form ethane:\[ \mathrm{C}_2\mathrm{H}_4(g) + \mathrm{H}_2(g) \rightarrow \mathrm{C}_2\mathrm{H}_6(g) \]The negative sign of \(\Delta V\) implies the gases contracted, so the system performed work on its surroundings. Calculating PV work yielded a negative work value, denoting an energy expenditure from the system to the environment. Thus, energy flowed outward. In endothermic reactions, energy is absorbed (energy flows into the system), while in exothermic ones, energy is released. Understanding this helps in assessing whether a reaction is consuming energy or releasing it, which can be critical in energy management and reaction design.