CBSE Class 11 Chemistry Thermodynamics Worksheet Set D - Free Printable

Let's solve each of these thermodynamics problems step by step.

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1. Which one of the following thermodynamic quantities is not a state function?

Options:
(a) Gibbs free energy
(b) enthalpy
(c) entropy
(d) internal energy
(e) work

Answer: (e) work

Explanation:
A state function depends only on the current state of the system, not on the path taken to reach that state. Examples include:
- Internal energy (U)
- Enthalpy (H)
- Entropy (S)
- Gibbs free energy (G)

Work, however, is not a state function because it depends on the path taken during a process (e.g., expansion vs. compression). Work is a path function.

✔ So, the correct answer is: (e) work

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2. A system has an increase in internal energy of 80 J and at the same time has 50 J of work done on it. What is the heat change of the system?

We use the First Law of Thermodynamics:

\[
\Delta U = q + w
\]

Where:
- \(\Delta U\) = change in internal energy
- \(q\) = heat added to the system
- \(w\) = work done on the system

Given:
- \(\Delta U = +80\, \text{J}\)
- \(w = +50\, \text{J}\) (since work is done on the system, it's positive)

Now solve for \(q\):

\[
80 = q + 50 \Rightarrow q = 80 - 50 = +30\, \text{J}
\]

So, 30 J of heat is added to the system.

✔ Answer: (b) +30 J

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3. The entropy will usually increase when:

I. A molecule is broken into two or more smaller molecules.
II. A reaction occurs that results in an increase in the number of moles of gas.
III. A solid changes to a liquid.
IV. A solid changes to a gas.

Options:
(a) I only
(b) II only
(c) III only
(d) IV only
(e) I, II, III, and IV

Explanation:
Entropy (\(S\)) is a measure of disorder or randomness. It increases when:
- Molecules become more disordered
- More freedom of motion
- Number of particles increases
- Phase changes from solid → liquid → gas

Let’s evaluate each:

- I. Breaking a molecule into smaller ones → more particles → more disorder → entropy increases ✔
- II. Increase in moles of gas → gases are highly disordered → entropy increases ✔
- III. Solid → liquid → more freedom of movement → entropy increases ✔
- IV. Solid → gas → huge increase in disorder → entropy increases ✔

All four scenarios lead to increased entropy.

✔ Answer: (e) I, II, III, and IV

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4. For the following reaction at 25°C, \(\Delta H^\circ = +115\, \text{kJ}\) and \(\Delta S^\circ = +125\, \text{J/K}\). Calculate \(\Delta G^\circ\) for the reaction at 25°C.

Reaction:
\[
\mathrm{SBr_4(g) \rightarrow S(s) + 2Br_2(l)}
\]

Use the equation:

\[
\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ
\]

Convert temperature to Kelvin:
\[
T = 25^\circ \text{C} = 298\, \text{K}
\]

Given:
- \(\Delta H^\circ = +115\, \text{kJ} = 115,000\, \text{J}\)
- \(\Delta S^\circ = +125\, \text{J/K}\)

Now calculate:

\[
\Delta G^\circ = 115,000\, \text{J} - (298\, \text{K})(125\, \text{J/K})
\]
\[
= 115,000 - 37,250 = 77,750\, \text{J} = 77.75\, \text{kJ}
\]

Rounded to one decimal place: +77.8 kJ

✔ Answer: (c) +77.8 kJ

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✔ Final Answers:

1. (e) work
2. (b) +30 J
3. (e) I, II, III, and IV
4. (c) +77.8 kJ

Let me know if you'd like explanations in simpler terms or diagrams!