# General Chemistry - Heat, Work, Enthelpy, Entropy Question

• modx07
In summary, the conversation discusses the calculation of q, w, ΔH, ΔS of the system, and ΔS of the surroundings when one mole of H2O(g) condenses to H2O(l) at 1.00atm and 95 Celcius. The equations used to calculate these values are q = nCΔT, ΔH = n(Cp)ΔT, W = -PΔV, and ΔS = [q_reversible] / T. The conversation also mentions the specific heat capacities of H2O(l) and H2O(g) and the ΔH-vaporization of H2O @ 100. The person asking the question is confused about
modx07
1. Assume that one mole of H2O(g) condenses to H2O(l) at 1.00atm and 95 Celcius. Calculate q, w, ΔH, ΔS of the system, ΔS of surroundings. BUT I AM NOT ASKING HOW TO CALCULATE THESE VALUES, SEE LAST SENTENCE OF POST.

## Homework Equations

q = nCΔT
ΔH = n(Cp)ΔT
W = -PΔV
ΔS = [q_reversible] / T

Cp of H2O(l) = 75.3 J / K*mol
Cp of H2O(g) = 36.4 J / K*mol
ΔH-vaporization of H2O @ 100 = 40.7 kJ / K*mol

## The Attempt at a Solution

Basically, I have the solution but I don't understand why it's the case. When I asked my TA, he said that the condensation is actually occurring at 95 Celcius, which confused me right off the bat since I knew that at 1atm, the condensation point should be 100 degrees.

Regardless, I thought that the step that was occurring should be simply:

H2O (g) -> H2O (l) @ 95 degrees.

Thus, I thought that since the system stays at the same temperature, then ΔE = 0 and that q = -w. However, this assumption is wrong. I am simply asking why this is wrong (and not to calculate the values).

Take this a point of discuss and not a solution, because I'm learning the same thing as we speak.

Since

ΔE = qv + w = nCvΔT + (-PextΔV)

and there is no change in volume, isn't it true that

ΔE = qv = nCvΔT

?

## 1. What is the difference between heat and work in chemistry?

Heat and work are both forms of energy transfer, but they differ in how they transfer energy. Heat is the transfer of energy due to a temperature difference, while work is the transfer of energy due to a force acting over a distance.

## 2. How is enthalpy related to chemical reactions?

Enthalpy is a measure of the total energy of a system, including both its internal energy and the work required to change its volume. In chemical reactions, enthalpy change (ΔH) is a measure of the heat released or absorbed during the reaction.

## 3. What is entropy and how does it relate to the second law of thermodynamics?

Entropy is a measure of the disorder or randomness of a system. The second law of thermodynamics states that the total entropy of a closed system (one that does not exchange matter or energy with its surroundings) will always increase over time. This means that energy tends to disperse and systems become more disordered over time.

## 4. How can I calculate the change in enthalpy and entropy for a chemical reaction?

The change in enthalpy (ΔH) for a reaction can be calculated by subtracting the enthalpies of the products from the enthalpies of the reactants. The change in entropy (ΔS) can be calculated by subtracting the entropies of the products from the entropies of the reactants. Both of these values can be found in thermodynamic tables or can be measured experimentally.

## 5. How do changes in temperature affect enthalpy and entropy in a chemical reaction?

Generally, an increase in temperature will increase both the enthalpy and entropy of a reaction. This is because higher temperatures allow for more energy to be transferred and for more disorder in the system. However, this relationship can vary depending on the specific reaction and conditions.

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