Assertion A: For an exothermic reaction at constant pressure, ΔH is negative. Reason R: Enthalpy is a state function. Choose the correct answer.

Both A and R are correct but R is not the correct explanation of A

A is correct but R is not correct

A is not correct but R is correct

Both A and R are correct and R is the correct explanation of A

Thermodynamics

11 NEET-UG practice questions on Thermodynamics, part of Chemistry. The first 11 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Medium
For a certain reaction, ΔH = −90 kJ/mol and ΔS = −200 J/K/mol, and neither changes with temperature. Under what condition can the reaction proceed spontaneously?
1. AAbove 450 K
2. BAt all temperatures
3. CBelow 450 K
4. DBelow 0.45 K
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Answer: C · Below 450 K
Spontaneity needs ΔG = ΔH − TΔS < 0. Here ΔG = −90000 + 200T, which is negative when T < 90000/200 = 450 K.
- A: This reverses the inequality; above 450 K, ΔG is positive.
- B: This holds only when ΔH is negative and ΔS is positive.
- D: This forgets to convert kJ to J.
Remember this
With ΔH and ΔS both negative, a reaction is spontaneous only below T = ΔH/ΔS.
Question 2 · Easy
During a process, a system absorbs 300 J of heat and does 120 J of work on its surroundings. What is the change in its internal energy?
1. A+420 J
2. B−420 J
3. C+180 J
4. D−180 J
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Answer: C · +180 J
Here q = +300 J, and work done by the system is w = −120 J. ΔU = q + w = 300 − 120 = +180 J.
- A: This takes work done by the system as positive.
- B: This takes both heat and work as losses.
- D: This reverses the sign of both q and w.
Remember this
In ΔU = q + w, heat absorbed and work done on the system are positive.
Question 3 · Medium
Assertion A: For an exothermic reaction at constant pressure, ΔH is negative. Reason R: Enthalpy is a state function. Choose the correct answer.
1. AA is correct but R is not correct
2. BA is not correct but R is correct
3. CBoth A and R are correct and R is the correct explanation of A
4. DBoth A and R are correct but R is not the correct explanation of A
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Answer: D · Both A and R are correct but R is not the correct explanation of A
A is true, as heat released at constant pressure equals ΔH, which is then negative. R is also true, since enthalpy depends only on the state of the system. But the sign of ΔH comes from heat being released, not from enthalpy being a state function.
- A: R is true, because enthalpy is a state function.
- B: A is true, because heat is released at constant pressure.
- C: R is true but does not account for the negative sign in A.
Remember this
A state function depends only on initial and final states; that fact does not fix the sign of ΔH.
Question 4 · Medium
Two moles of an ideal gas expand isothermally from 1 L to 6 L against a constant external pressure of 2 bar. How much work is done on the gas? (Take 1 L bar = 100 J)
1. A−1200 J
2. B−10 J
3. C+1000 J
4. D−1000 J
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Answer: D · −1000 J
w = −P(ext)ΔV = −2 × (6 − 1) = −10 L bar. Converting, −10 × 100 = −1000 J. The number of moles is not needed for an expansion against constant pressure.
- A: This uses the final volume rather than the change in volume.
- B: This leaves the answer in L bar and writes it as joules.
- C: This misses the negative sign for expansion.
Remember this
Against constant external pressure, w = −P(ext)ΔV, and it is negative for expansion.
Question 5 · Medium
Match List-I with List-II. List-I (signs for a process) A. ΔH < 0, ΔS < 0 B. ΔH > 0, ΔS > 0 C. ΔH < 0, ΔS > 0 D. ΔH > 0, ΔS < 0 List-II (spontaneity) I. Spontaneous at all temperatures II. Non-spontaneous at all temperatures III. Spontaneous only at low temperature IV. Spontaneous only at high temperature
1. AA-III, B-IV, C-I, D-II
2. BA-III, B-IV, C-II, D-I
3. CA-IV, B-III, C-I, D-II
4. DA-I, B-II, C-III, D-IV
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Answer: A · A-III, B-IV, C-I, D-II
ΔG = ΔH − TΔS. In A, the −TΔS term is positive and small only at low T. In B, the −TΔS term is negative and large only at high T. In C both terms are negative and in D both are positive.
- B: This swaps the always spontaneous and never spontaneous cases.
- C: This swaps the low and high temperature cases.
- D: This matches the lists in order without using ΔG = ΔH − TΔS.
Remember this
Signs of ΔH and ΔS that agree in favouring spontaneity make it hold at all temperatures; those that disagree make it depend on T.
Question 6 · Medium
At 298 K, the standard enthalpies of formation are CO₂(g) = −394, H₂O(l) = −286 and C₂H₅OH(l) = −278 kJ/mol. Calculate the standard enthalpy of combustion of liquid ethanol.
1. A−1924 kJ/mol
2. B+1368 kJ/mol
3. C−1646 kJ/mol
4. D−1368 kJ/mol
Answer and explanation in LearnFree account
Question 7 · Easy
An ideal gas expands isothermally into a vacuum. Which set of values holds for this process?
1. Aq = 0, w < 0, ΔU < 0
2. Bq > 0, w = 0, ΔU > 0
3. Cq = 0, w = 0, ΔU = 0
4. Dq > 0, w < 0, ΔU = 0
Answer and explanation in LearnFree account
Question 8 · Hard
Methane (1.6 g) is burned completely in a bomb calorimeter of heat capacity 22 kJ/K, and the temperature rises by 4 K at about 300 K: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) Find ΔH for this reaction per mole of CH₄. (C = 12, H = 1; take R = 8 J/K/mol)
1. A−875.2 kJ/mol
2. B−880.0 kJ/mol
3. C−882.4 kJ/mol
4. D−884.8 kJ/mol
Answer and explanation in LearnFree account
Question 9 · Hard
In the enthalpy diagram all values are per mole of reaction at 298 K. What is the standard enthalpy of formation of methane, marked X?
Figure: MedExams
1. A−75 kJ mol⁻¹
2. B+75 kJ mol⁻¹
3. C−965 kJ mol⁻¹
4. D−1855 kJ mol⁻¹
Answer and explanation in LearnFree account
Question 10 · Easy
An ideal gas expands along the path shown, against a constant external pressure equal to the gas pressure. How much work is done on the gas?
Figure: MedExams
1. A−6 L bar
2. B−18 L bar
3. C−12 L bar
4. D+12 L bar
Answer and explanation in LearnFree account
Question 11 · Medium
ΔG of a process changes with temperature as shown, with ΔH and ΔS constant over the range. What is ΔS of the process?
Figure: MedExams
1. A+0.1 J K⁻¹ mol⁻¹
2. B+100 J K⁻¹ mol⁻¹
3. C−100 J K⁻¹ mol⁻¹
4. D+67 J K⁻¹ mol⁻¹
Answer and explanation in LearnFree account

Thermodynamics

For a certain reaction, ΔH = −90 kJ/mol and ΔS = −200 J/K/mol, and neither changes with temperature. Under what condition can the reaction proceed spontaneously?

During a process, a system absorbs 300 J of heat and does 120 J of work on its surroundings. What is the change in its internal energy?

Assertion A: For an exothermic reaction at constant pressure, ΔH is negative. Reason R: Enthalpy is a state function. Choose the correct answer.

Two moles of an ideal gas expand isothermally from 1 L to 6 L against a constant external pressure of 2 bar. How much work is done on the gas? (Take 1 L bar = 100 J)

At 298 K, the standard enthalpies of formation are CO₂(g) = −394, H₂O(l) = −286 and C₂H₅OH(l) = −278 kJ/mol. Calculate the standard enthalpy of combustion of liquid ethanol.

An ideal gas expands isothermally into a vacuum. Which set of values holds for this process?

Methane (1.6 g) is burned completely in a bomb calorimeter of heat capacity 22 kJ/K, and the temperature rises by 4 K at about 300 K: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) Find ΔH for this reaction per mole of CH₄. (C = 12, H = 1; take R = 8 J/K/mol)

In the enthalpy diagram all values are per mole of reaction at 298 K. What is the standard enthalpy of formation of methane, marked X?

An ideal gas expands along the path shown, against a constant external pressure equal to the gas pressure. How much work is done on the gas?

ΔG of a process changes with temperature as shown, with ΔH and ΔS constant over the range. What is ΔS of the process?