Question

Consider the reaction between methane and oxygen producing carbon dioxide and water. Suppose that the reaction is carried out in a furnace used to heat a house. If \(q=-890 \mathrm{~kJ}\) and \(w=+5 \mathrm{~kJ},\) what is \(\Delta E ? \Delta H\) at \(25^{\circ} \mathrm{C} ?\)

Short answer

Answer: The change in internal energy (ΔE) is -885 kJ, and the expression for the change in enthalpy (ΔH) is ΔH = -885 kJ + PΔV.

Step-by-step solution

Calculate ΔE

To find the change in internal energy (ΔE), we will use the first law of thermodynamics: ΔE = q + w. We are given the values of heat transfer (q) and work (w) as -890 kJ and +5 kJ, respectively. Now substitute the values in the formula: ΔE = (-890 kJ) + (+5 kJ) ΔE = -885 kJ

Calculate ΔH

Under constant pressure (which is the case here), the relationship between ΔH, ΔE, and PΔV is given by: ΔH = ΔE + PΔV However, since we are not given any information about the pressure or the change in volume of the system, we cannot determine the numerical value of ΔH for this process. We can only express it in terms of ΔE and PΔV: ΔH = -885 kJ + PΔV To find the exact value of ΔH, we would need more information about the pressure and the change in volume of the system during the reaction.

Key concepts

First Law of Thermodynamics

The first law of thermodynamics is a version of the law of conservation of energy tailored for thermodynamic systems. It asserts that energy cannot be created or destroyed in an isolated system. Instead, energy can only be transferred or changed from one form to another. This principle is fundamental in studying reactions such as the combustion of methane to heat a house.

In the context of our exercise, the first law is expressed as \( \Delta E = q + w \), where \( \Delta E \) represents the change in internal energy of the system, \( q \) is the heat exchanged with the surroundings, and \( w \) is the work done by or on the system. When methane burns, -890 kJ of heat is released (as indicated by \( q = -890 \: \text{kJ} \) meaning the system lost heat), and the system does work on the surroundings (here the furnace pushing out exhaust gases) by +5 kJ. Hence, applying the first law to our problem gives us the change in internal energy for the reaction.

Enthalpy Change

Enthalpy change (\(\Delta H\)) is another crucial concept in thermochemistry, particularly when reactions occur at constant pressure, such as in the open atmosphere or in a furnace vented to the outside. It reflects the heat absorbed or released by the system. Enthalpy change is given by the relationship \( \Delta H = \Delta E + P\Delta V \), where \( P\Delta V \) accounts for the work done against atmospheric pressure as the system expands or contracts.

In our exercise, we do not have enough information to calculate the exact value of \( \Delta H \) because the pressure and volume change \( P\Delta V \) are not provided. It is essential to understand that \( \Delta H \) is a state function, meaning it depends only on the initial and final states of the system, not the path taken. Hence, any attempt to calculate the enthalpy change would require details about the system's initial and final conditions regarding pressure and volume.

Internal Energy

Internal energy is the total energy contained within a system and is comprised of the potential and kinetic energy of all the particles within the system. In a chemical reaction such as the combustion of methane, internal energy changes reflect the breaking and forming of chemical bonds.

According to the problem, using the provided values of \( q \) and \( w \) in the expression of the first law, we calculated the change in internal energy \( \Delta E = -885 \: \text{kJ} \) for the system. It's crucial to understand the significance of the sign: a negative \( \Delta E \) indicates that the system released energy to the surroundings, consistent with the exothermic nature of a combustion reaction.