NEET (UG)

Practice Test 1 — Thermodynamics

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Section A — MCQ (Single Correct & Statement-based)

Question 1

The first law of thermodynamics is written as:

Solution: $\Delta U = q + W$ with work-on-system convention.

Question 2

Which quantity is NOT a state function?

Solution: Work is a path function.

Question 3

At constant pressure, $q_p$ equals:

Solution: $q_p = \Delta H$.

Question 4

Statements: (I) Exothermic reactions have $\Delta H<0$. (II) Hess's law follows from enthalpy being a state function. Which are correct?

Solution: Both statements are correct.

Question 5

For $\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$, $\Delta n_g$ is:

Solution: $0 - 1.5 = -1.5$ (liquid product has no gas moles).

Question 6

Entropy increases most in which change?

Solution: Solid to gas is the largest disorder increase.

Question 7

A reaction with $\Delta H>0$ and $\Delta S<0$ is:

Solution: $\Delta G > 0$ at all T.

Question 8

If $\Delta G^{\circ}$ is large and negative, the equilibrium constant $K$ is:

Solution: $\Delta G^{\circ} = -RT\ln K$; negative $\Delta G^{\circ}$ to large $K$.

Section B — Assertion & Reason

Question 9

A: Work done is not a state function.
R: The work done in a process depends on the path taken between the initial and final states.

Solution: Path-dependence is exactly why work is not a state function — R explains A.

Question 10

A: An endothermic reaction can be spontaneous.
R: A sufficiently large positive $\Delta S$ can make $\Delta G$ negative even when $\Delta H$ is positive.

Solution: A large $T\Delta S$ can overcome a positive $\Delta H$ — R explains A.

Question 11

A: The entropy of the system always increases in a spontaneous process.
R: The second law requires the entropy of the universe to increase.

Solution: The system's entropy can decrease (e.g. freezing) as long as the universe's rises, so A is false; R correctly states the second law.

Question 12

A: At equilibrium the Gibbs free energy change $\Delta G$ is zero.
R: At equilibrium there is no net tendency for the reaction to proceed in either direction.

Solution: No net drive corresponds to $\Delta G = 0$ — R explains A.