Problem 35 - Chemistry: The Central Science in SI Units, Global Edition

Question

Assume that 2 moles of water are formed according to the following reaction at constant pressure \((101.3 \mathrm{kPa})\) and constant temnerature \((298 \mathrm{~K});\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the pressure-volume work for this reaction. (b) Calculate \(\Delta E\) for the reaction using your answer to (a).

Step-by-Step Solution

Answer

(a) The pressure-volume work, W, for the reaction is calculated as 7.47 kJ. (b) The change in internal energy, ΔE, for the reaction is -1135.73 kJ.

Step 1: Finding the volume change in the reaction

First, we need to find the initial and final moles of gas in the reaction. Initially, we have 2 moles of hydrogen gas and 1 mole of oxygen gas. So, initial moles of gas: \(n_{initial} = 2 + 1 = 3\) moles After the reaction, 2 moles of water (liquid) are formed, and there are no moles of gas left in the reaction. So, final moles of gas: \(n_{final} = 0\) moles Now, we'll use the Ideal Gas Law, with the given temperature \(T = 298 \mathrm{K}\) and pressure \(P = 101.3\,\mathrm{kPa}\), to find the initial and final volumes: $$PV = nRT$$ Where \(R\) is the gas constant in \(\mathrm{kPa \cdot L/mol\cdot K}\) which is equal to $8.314 \,\mathrm{J/mol\cdot K} \div 1000 = 0.008314\,\mathrm{kPa\cdot L/mol\cdot K}\) Initial volume: $$V_{initial} = \frac{n_{initial}RT}{P} = \frac{3(0.008314)(298)}{101.3} = 0.0737 \, \mathrm{L}$$ Final volume (V_final) is 0 since there are no moles of gas left in the reaction \((n_{final} = 0)\). The volume change (ΔV) during the reaction: $$\Delta V = V_{final} - V_{initial} = 0 - 0.0737 = -0.0737 \, \mathrm{L}$$

Step 2: Calculate the pressure-volume work

The pressure-volume work for a constant pressure process is given by: $$W = -P\Delta V$$ Now, plugging the given pressure and calculated volume change: $$W = -(101.3)(-0.0737) = 7.47 \, \mathrm{kJ}$$ So, the pressure-volume work for this reaction is 7.47 kJ.

Step 3: Calculate the change in internal energy

Since we're given the pressure-volume work (W) and asked to find the change in internal energy (ΔE), we can use the following equation: $$ΔE = q + W$$ However, we're not given the heat transferred (q) directly. But we can use the following fact: for any reaction at constant temperature and pressure, the heat transferred is equivalent to the change in enthalpy (ΔH) of the reaction. For the formation of 2 moles of water from hydrogen and oxygen gases, the change in enthalpy is: $$ΔH = -571.6 \, \mathrm{kJ/mol}$$ Since 2 moles of water are formed, the total change in enthalpy is: $$q = 2 \times (-571.6) = -1143.2 \, \mathrm{kJ}$$ Now we can find the change in internal energy using the relation: $$ΔE = q + W = -1143.2 + 7.47 = -1135.73 \, \mathrm{kJ}$$ Therefore, the change in internal energy (ΔE) for the reaction is -1135.73 kJ.

Key Concepts

Enthalpy

In thermochemistry, enthalpy (\( H \)) represents the total heat content of a system. This is especially important in chemical reactions at constant pressure. Enthalpy change (\( \Delta H \)) is used to express the heat absorbed or released by a reaction.
For reactions, the change in enthalpy is given by:

• Positive \( \Delta H \): the system absorbs energy (endothermic reaction).
• Negative \( \Delta H \): the system releases energy (exothermic reaction).

In our provided reaction, where two moles of hydrogen and one mole of oxygen form two moles of liquid water, the enthalpy change is \( \Delta H = -1143.2 \, \text{kJ} \).
A negative value indicates that the reaction releases energy, or in other words, it's exothermic. This loss of energy as heat is a key feature of this type of reaction, forming stronger bonds in liquid water than the initial gaseous states of hydrogen and oxygen.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and the number of moles of a gas. The formula is given by:
\[ PV = nRT \]
where \( P \) is pressure in kilopascals, \( V \) is volume in liters, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.008314 \, \text{kPa} \cdot \text{L/mol} \cdot \text{K}), and \( T \) is temperature in Kelvin.
This equation is used to calculate the volume of a gas under certain conditions. In the exercise, the Ideal Gas Law helps determine the initial volume change due to the consumption of gases in the reaction.
The initial scenario has 3 moles of gas, whereas the resulting scenario has 0 moles. This drop from 3 to 0 moles illustrates a significant volume change, allowing us to calculate the work done during this chemical reaction when pressure remains constant.

Internal Energy

Internal Energy (\( E \)) in a chemical reaction refers to the total energy contained within the system. It includes all forms of energy the molecules possess within a system. The change in internal energy (\( \Delta E \)) during a chemical process is a crucial step to understand energy conservation. It is calculated by:
\[ \Delta E = q + W \]
where \( q \) represents heat exchange, and \( W \) is the work done during the reaction.
In reactions at constant pressure, where temperature remains consistent, the heat component (\( q \)) equates to the enthalpy change (\( \Delta H \)).
In our case, the internal energy change derived from the step-by-step process is \( \Delta E = -1135.73 \, \text{kJ} \).
This calculation indicates the net energy loss within the system, emphasizing how energy is conserved throughout chemical transformations. In this exercise, it's evident that the reaction's energy loss is significantly due to exiting heat, affirming the principle of energy conservation.