Question

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J} ;(\mathbf{b})\) 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; (c) the system absorbs \(7.25 \mathrm{~kJ}\) of heat from the surroundings while its volume remains constant (assume that only \(P-V\) work can be done).

Short answer

For the given cases, we have calculated the following values for \(\Delta E\): - Case (a): \(\Delta E = -77 \mathrm{~J}\), indicating an exothermic process. - Case (b): \(\Delta E = -22100 \mathrm{~J}\), indicating an exothermic process. - Case (c): \(\Delta E = 7250 \mathrm{~J}\), indicating an endothermic process.

Step-by-step solution

Case (a)

Calculate \(\Delta E\) and determine if the process is endothermic or exothermic. We have q = 0.763 kJ and w = -840 J. First, we need to convert q into joules: 0.763 kJ * 1000 J/kJ = 763 J Now we can use the formula: \[\Delta E = q + w = 763 \mathrm{~J} + (-840 \mathrm{~J}) = -77 \mathrm{~J}\] Since \(\Delta E\) is negative, the process is exothermic, which means it releases heat.

Case (b)

First, let's establish the signs of q and w. Since the system releases heat, q is negative. Since the surroundings do work on the system, w is positive. Now, convert the given kJ to J: q = -66.1 kJ * 1000 J/kJ = -66100 J w = 44.0 kJ * 1000 J/kJ = 44000 J Next, we'll calculate \(\Delta E\): \[\Delta E = q + w = (-66100 \mathrm{~J}) + (44000 \mathrm{~J}) = -22100 \mathrm{~J}\] With a negative \(\Delta E\), the process is exothermic, meaning it releases heat.

Case (c)

Since the volume remains constant and only \(P-V\) work can be done, there is no work being done in this case (w = 0). The system absorbs heat from the surrounding so q is positive: q = 7.25 kJ * 1000 J/kJ = 7250 J Now compute \(\Delta E\): \[\Delta E = q + w = 7250 \mathrm{~J} + 0 = 7250 \mathrm{~J}\] Since \(\Delta E\) is positive, the process is endothermic, which means it absorbs heat.

Key concepts

Endothermic Processes

Endothermic processes are fascinating. In these processes, a system absorbs energy from its surroundings. This energy is usually in the form of heat.In case (c) from our original exercise, we encountered an endothermic reaction. Here, the system absorbed heat, with a calculated energy change (\(\Delta E = 7250 \mathrm{~J}\)). This positive energy change indicates that the system gained energy. With endothermic reactions, temperature changes can often feel cold to the touch, as heat is absorbed rather than released.Let's briefly explore some common endothermic processes:

• Melting ice.
• Evaporating water.
• Photosynthesis in plants.

In many of these processes, the system (be it ice, water, or a plant) requires energy to move to a higher energy state. Endothermic reactions are essential in both natural cycles and industrial processes.

Exothermic Processes

Exothermic processes release energy to the surroundings. This type of energy release typically involves heat, resulting in an increase in temperature around the system.Examining cases (a) and (b) from our exercise helps emphasize this concept. Both instances resulted in a negative \(\Delta E\), indicating a release of energy:- Case (a) had \(\Delta E = -77 \mathrm{~J}\).- Case (b) had \(\Delta E = -22100 \mathrm{~J}\).Thermochemically, this loss of energy is felt as heat generation. Exothermic reactions are everywhere:

• Combustion, like burning wood or gasoline.
• Respiration in living cells.
• Mixing strong acids and bases.

In these reactions, energy is released when bonds form in reaction products. Exothermic reactions are vital in many applications, such as engine fuels and metabolic processes in organisms.

Energy Change Calculations

Calculating energy changes requires a clear understanding of the relationship between heat (\(q\)) and work (\(w\)). This relationship is expressed as:\[ \Delta E = q + w \]Let's detail this formula using our exercise:

• Case (a): Converted heat and work gave us \(\Delta E = -77 \mathrm{~J}\).
• Case (b): Converted measurements resulted in \(\Delta E = -22100 \mathrm{~J}\).
• Case (c): With a constant volume, w = 0, leaving \(\Delta E = 7250 \mathrm{~J}\) solely from the absorbed heat.

Notice the need to convert kilojoules to joules. This step ensures consistency in units, making the calculations yielding reliable results easier. Understanding \(\Delta E\) helps predict whether a reaction absorbs or releases energy. This understanding is foundational in thermochemistry, directly impacting how we control reactions in lab and industrial settings.Precision in energy change calculations not only determines heating and cooling requirements but also ensures efficiency in energy usage across various processes.