Class 11 Chemistry · Original EZT practice
Thermodynamics (Class 11 Chapter 5): summary and practice
Thermodynamics is the bookkeeping of energy: what the system gains as heat or work, how enthalpy packages it at constant pressure, and how Gibbs energy decides whether a reaction runs on its own. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Thermodynamic terms
The system is the part under study and everything else is the surroundings; systems are open, closed or isolated. State functions like internal energy U depend only on the state, not the path taken to reach it.
2. First law and sign convention
The chapter writes the first law as delta U = q + w, where q is heat absorbed by the system and w is work done on the system. Expansion work against external pressure is w = -p_ex delta V, so work done by the system is negative.
3. Enthalpy
H = U + pV. At constant pressure the chapter shows delta H = q_p, and for reactions involving gases delta H = delta U + delta n_g RT, where delta n_g counts the change in moles of gas.
4. Measuring delta U and delta H
A bomb calorimeter at constant volume measures q_v = delta U; a coffee-cup calorimeter at constant pressure measures q_p = delta H. Heat capacity links the temperature rise to the heat.
5. Reaction enthalpies and Hess's law
Standard reaction enthalpy can be built from standard enthalpies of formation, combustion or other data. Hess's law: the total enthalpy change is the same whether the reaction happens in one step or several.
6. Spontaneity and Gibbs energy
Enthalpy alone does not decide spontaneity - entropy matters too. Gibbs energy combines them: delta G = delta H - T delta S, and a process is spontaneous at constant T and p when delta G is negative.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Apply the first law with signs
Choose the method: delta U = q + w, with w positive when work is done on the system.
EZT practice question
A system absorbs 500 J of heat and does 200 J of work on the surroundings. Find delta U.
Show the worked solution
- q = +500 J (heat absorbed).
- w = -200 J (work done by the system).
- delta U = q + w = 500 - 200.
Answer: delta U = +300 J.
Check: Absorbed heat adds energy; work done by the system removes it.
Watch for this: Sign of w flips the answer - the chapter's convention is work done on the system.
Now try it yourself
A system releases 150 J of heat and 60 J of work is done on it - delta U?
Check your answer
-90 J.
q = -150 J, w = +60 J, so delta U = -90 J.
Convert between delta H and delta U
Choose the method: delta H = delta U + delta n_g RT.
EZT practice question
For N2 + 3H2 -> 2NH3 at 298 K, delta U = -90.0 kJ. Find delta H (R = 8.314 J/K mol).
Show the worked solution
- delta n_g = 2 - (1 + 3) = -2.
- delta n_g RT = -2 x 8.314 x 298 = -4955 J = -4.955 kJ.
- delta H = -90.0 + (-4.955).
Answer: delta H = -94.96 kJ.
Check: Fewer gas moles on the product side makes delta H more negative than delta U here.
Watch for this: Convert J to kJ before adding - mixing the units is the classic error.
Now try it yourself
A reaction has delta n_g = +1 at 300 K - how much do delta H and delta U differ?
Check your answer
2.49 kJ.
delta n_g RT = 1 x 8.314 x 300 = 2494 J = 2.49 kJ.
Use Hess's law
Choose the method: Add the step enthalpies along any path.
EZT practice question
A -> B has delta H = -100 kJ and B -> C has delta H = +30 kJ. Find delta H for A -> C.
Show the worked solution
- Hess's law: the total change is path-independent.
- delta H(A -> C) = -100 + 30.
Answer: delta H = -70 kJ.
Check: The two-step path and the direct path must give the same net change.
Watch for this: Watch the signs when a step is reversed - reversing a reaction flips the sign of its delta H.
Now try it yourself
A -> B is -100 kJ; what is delta H for B -> A?
Check your answer
+100 kJ.
Reversing a reaction flips the sign.
Judge spontaneity with Gibbs energy
Choose the method: delta G = delta H - T delta S; spontaneous when delta G < 0.
EZT practice question
For a reaction, delta H = -92 kJ and delta S = -199 J/K. Is it spontaneous at 298 K?
Show the worked solution
- Convert delta S to kJ: -0.199 kJ/K.
- T delta S = 298 x (-0.199) = -59.30 kJ.
- delta G = -92 - (-59.30).
Answer: delta G = -32.70 kJ, so the reaction is spontaneous at 298 K.
Check: A negative delta H can outweigh a negative delta S at low enough temperature.
Watch for this: delta S is usually in J/K while delta H is in kJ - convert before combining.
Now try it yourself
Same reaction at 500 K - delta G?
Check your answer
+7.5 kJ, not spontaneous.
T delta S = 500 x (-0.199) = -99.5 kJ, so delta G = -92 + 99.5 = +7.5 kJ.
Match the calorimeter to the quantity
Choose the method: Constant volume gives delta U; constant pressure gives delta H.
EZT practice question
Which quantity does a bomb calorimeter measure directly, and why?
Show the worked solution
- A bomb calorimeter runs at constant volume.
- At constant volume no expansion work is done, so the heat released equals delta U (q_v = delta U).
Answer: It measures delta U, because constant volume means q_v = delta U.
Check: A coffee-cup calorimeter at constant pressure instead measures q_p = delta H.
Watch for this: Do not quote delta H from a bomb calorimeter without converting through delta n_g RT.
Now try it yourself
Which calorimeter measures delta H directly?
Check your answer
The constant-pressure (coffee-cup) calorimeter.
At constant pressure, q_p = delta H.
Identify state functions
Choose the method: State functions depend only on the state, not the path.
EZT practice question
Is work a state function? Explain with the chapter's reasoning.
Show the worked solution
- Internal energy U is a state function: its change is fixed by the end states.
- Heat q and work w depend on the path taken, so they are not state functions - only their sum delta U is path-independent.
Answer: No - work depends on the path, so it is not a state function.
Check: This is exactly why the first law matters: two path-dependent quantities add to a path-independent one.
Watch for this: Do not call q or w state functions even though delta U = q + w is one.
Now try it yourself
Is enthalpy H a state function?
Check your answer
Yes.
H = U + pV is built from state functions, so it is one too.
Try a mixed set
Choose a method or explanation before opening each answer.
1. First law as the chapter writes it?
Show answer and reason
delta U = q + w.
w is work done on the system.
2. delta H from delta U?
Show answer and reason
delta H = delta U + delta n_g RT.
For reactions involving gases.
3. Gibbs energy relation?
Show answer and reason
delta G = delta H - T delta S.
Negative delta G means spontaneous at constant T and p.
4. Hess's law in one line?
Show answer and reason
Total enthalpy change is the same for any path.
Enthalpy is a state function.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A system releases 150 J of heat and has 60 J of work done on it: find delta U.
- A reaction has delta n_g = +1 at 300 K: by how much do delta H and delta U differ?
- delta H = -92 kJ, delta S = -199 J/K: is the reaction spontaneous at 500 K?
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Questions about this practice page
- What is the difference between a state function and a path function?
- A state function (U, H, S, G) depends only on the current state of the system. Path functions like heat q and work w depend on how the change is carried out.
- Why does delta H equal q at constant pressure?
- From H = U + pV, at constant pressure delta H = delta U + p delta V. Substituting the first law with expansion work gives delta H = q_p, as the chapter derives.
- When is a reaction spontaneous?
- When delta G = delta H - T delta S is negative at constant temperature and pressure. Exothermic reactions with positive entropy are spontaneous at all temperatures; other combinations depend on T.
- What does Hess's law let you do?
- Add or subtract known reaction enthalpies along any path to get the enthalpy change of a reaction you cannot measure directly.
- Are these the NCERT exercise questions for Thermodynamics (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.