Question

Calculate $\mathrm{\Delta }E$ and determine whether the process is endothermic or exothermic for the following cases: $(\mathbf{a})q=0.763\mathrm{kJ}$ and $w=-840\mathrm{J}$. (b) A system releases $66.1\mathrm{kJ}$ of heat to its surroundings while the surroundings do $44.0\mathrm{kJ}$ of work on the system.

Short answer

In case (a): ΔE = -0.077 kJ, making the process exothermic. In case (b): ΔE = -22.1 kJ, also making the process exothermic.

Step-by-step solution

Calculate ΔE using ΔE = q + w

For this case, we're given values for q and w. We just need to plug these values into the equation: ΔE = q + w. Note that q and w have different units, so we need to make sure they have the same units before addition. w is given in Joules (J), while q is given in kilojoules (kJ). We can convert w to kJ by dividing it by 1000. So, ΔE = (0.763 kJ) + (-840 J * 1 kJ/1000 J) ΔE = 0.763 kJ - 0.84 kJ

Determine whether the process is endothermic or exothermic

Now that the units are the same, we can calculate ΔE: ΔE = -0.077 kJ Since ΔE is negative, the process is exothermic. Case (b):

Determine the signs of q and w

We're given that the system releases 66.1 kJ of heat to its surroundings, which means q = -66.1 kJ (heat released is negative). We're also given that the surroundings do 44.0 kJ of work on the system, which means w = +44.0 kJ (work done on the system is positive).

Calculate ΔE using ΔE = q + w

With the signs of q and w determined, we can calculate ΔE: ΔE = q + w ΔE = (-66.1 kJ) + (44.0 kJ)

Determine whether the process is endothermic or exothermic

Now we can calculate ΔE: ΔE = -22.1 kJ Since ΔE is negative, the process is exothermic.

Key concepts

Endothermic Processes

In thermochemistry, endothermic processes are characterized by the absorption of heat from the surroundings into the system. During such processes, the system gains energy, leading to an increase in internal energy, often reflected by a positive change in enthalpy $\mathrm{\Delta }H$.
To identify an endothermic reaction, look for cues such as:

• Heat being added to the system
• A positive value for heat change ($q>0$)
• The process feeling cold to the touch as heat is absorbed from the environment

For example, melting ice is an endothermic process because it requires heat to change from a solid to a liquid form. The key indicator of endothermicity is the sign of the energy change, where $\mathrm{\Delta }E$ or the enthalpy change is positive.

Exothermic Processes

Exothermic processes occur when a system releases heat to its surroundings. In these processes, the system loses energy, which is manifesting as a negative change in internal energy $\mathrm{\Delta }E$ or enthalpy $\mathrm{\Delta }H$.
An easy way to recognize exothermic reactions is by their outcomes:

• There is a release of heat, resulting in the surroundings becoming warmer
• The change in energy (q) is negative, indicating heat being released ($q<0$)
• These processes often feel hot to the touch

A classic example of an exothermic process is the combustion of gasoline, where heat is released, powering engines and producing an increase in temperature around the reaction zone. It's important in thermochemical calculations to determine whether a process is exothermic or endothermic by examining the sign of $\mathrm{\Delta }E$. A negative $\mathrm{\Delta }E$ implies energy is released, confirming the process is exothermic.

Energy Change Calculation

Calculating the change in internal energy $\mathrm{\Delta }E$ of a system is fundamental in thermochemistry. The formula to find $\mathrm{\Delta }E$ is $‘\mathrm{\Delta }E=q+w‘$, where:

• $q$ represents the heat exchanged with the surroundings
• $w$ is the work done on or by the system

To perform the calculation, ensure both q and w are in the same units. Most commonly, $q$ and $w$ are measured in joules (J) or kilojoules (kJ). If needed, convert the units so they match.
For instance, solving the energy change from our examples:
1. For the first case, the conversion is necessary because $w$ is in joules and needs to be converted to kilojoules before combining it with $q$ using the equation:
$$\mathrm{\Delta }E=(0.763\text{ kJ})+(-840\text{ J}\times \frac{1\text{ kJ}}{1000\text{ J}})=0.763\text{ kJ}-0.84\text{ kJ}$$
This results in $\mathrm{\Delta }E=-0.077$ kJ, and since it's negative, the process is exothermic.
2. Similarly, for the second case:
$$\mathrm{\Delta }E=(-66.1\text{ kJ})+(44.0\text{ kJ})=-22.1\text{ kJ}$$
Again, the negative result affirms that this is an exothermic process. The computation of $\mathrm{\Delta }E$ not only reveals the direction of energy flow but also helps in understanding the nature of the process involved.