Problem 110 - Chemistry and Chemical Reactivity

Question

In the reaction of two moles of gaseous hydrogen and one mole of gaseous oxygen to form two moles of gaseous water vapor, two moles of products are formed from 3 moles of reactants. If this reaction is done at \(\left.1.0 \text { atm pressure (and at } 0^{\circ} \mathrm{C}\right),\) the volume is reduced by \(22.4 \mathrm{L}\) (a) In this reaction, how much work is done on the system \(\left(\mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O}\right)\) by the surroundings? (b) The enthalpy change for this reaction is \(-483.6 \mathrm{kJ}\) Use this value, along with the answer to (a), to calculate \(\Delta_{r} U\), the change in internal energy in the system.

Step-by-Step Solution

Answer

(a) 2.27072 kJ work is done on the system. (b) \( \Delta U = -485.87072 \text{ kJ} \).

Step 1: Understand the Reaction

The reaction given is: \( 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \). You have 3 moles of gaseous reactants that combine to form 2 moles of gaseous products at constant conditions of 1 atm and 0°C. This results in a volume decrease of 22.4 L.

Step 2: Calculate Work Done by the Surroundings

Work done \( w \) on a system by the surroundings can be calculated using the formula: \( w = -P\Delta V \). Here, \( P = 1 \text{ atm } = 101.3 \text{ J/L} \) and \( \Delta V = -22.4 \text{ L} \) (since volume decreases). Thus, \( w = -1 \text{ atm} \times (-22.4 \text{ L}) \times (101.3 \text{ J/L}) = 2270.72 \text{ J} = 2.27072 \text{ kJ} \). Hence, the work done on the system is \( 2.27072 \text{ kJ} \).

Step 3: Use Enthalpy to Determine Internal Energy Change

The change in internal energy \( \Delta U \) can be found using the relation: \( \Delta U = \Delta H - w \). Given \( \Delta H = -483.6 \text{ kJ} \) and the work done \( w = 2.27072 \text{ kJ} \) (remembering work done *on* the system is positive), substitute these values: \( \Delta U = -483.6 \text{ kJ} - 2.27072 \text{ kJ} = -485.87072 \text{ kJ} \).

Key Concepts

Enthalpy

Enthalpy, denoted as \( \Delta H \), quantifies the total heat content of a system. It is especially useful in thermochemistry as it describes the heat change at constant pressure. The reaction provided has an enthalpy change of \( -483.6 \text{ kJ} \). This negative value indicates that the reaction releases energy to the surroundings, known as an exothermic reaction.

In chemical processes, enthalpy changes give us insight into whether energy is absorbed or released.

• Exothermic reactions have negative \( \Delta H \) values, releasing heat.
• Endothermic reactions have positive \( \Delta H \) values, absorbing heat.

For this specific reaction forming water vapor from hydrogen and oxygen, the enthalpy change implies that the system releases 483.6 kJ of energy. This energy release helps to drive the formation of the product and stabilize it energetically.

Internal Energy Change

Internal energy change, represented as \( \Delta U \), varies differently from enthalpy as it includes all energy changes within a system, not just the heat at constant pressure. To find \( \Delta U \), we use the equation:\[\Delta U = \Delta H - w\]where \( w \) is the work done on or by the system.

In this case:

• \( \Delta H = -483.6 \text{ kJ} \)
• \( w = 2.27072 \text{ kJ} \)

Substituting these values, we get:\[\Delta U = -483.6 \text{ kJ} - 2.27072 \text{ kJ} = -485.87072 \text{ kJ}\]This calculation shows that the total internal energy of the system decreases. The energy leaving the system indicates that a greater amount of energy is being expended due to both heat loss and work being performed.

Work Done on the System

Work done on a system is represented by the work (\( w \)) formula:\[w = -P\Delta V\]where

• \( P \) stands for pressure, equivalent to 1 atm or 101.3 J/L under standard conditions
• \( \Delta V \) is the change in volume, equating to \(-22.4 \text{ L} \) for this reaction because the volume decreases.

Thus, calculating work gives us:\[w = -1 \text{ atm} \times (-22.4 \text{ L}) \times (101.3 \text{ J/L}) = 2270.72 \text{ J} = 2.27072 \text{ kJ}\]This calculation shows 2.27072 kJ of work is done on the system.

Total work is positive when it's done on the system and negative if work is done by the system. Through these calculations, we see that the surrounding pressure performs work on the chemical system as it contracts, contributing to the overall energy dynamics between the reactants and products.