Question 1

The first law of thermodynamics states that ΔU = q + w. What does 'w' represent?

Accepted Answer

In the equation ΔU = q + w, 'w' represents work done ON the system (positive when work is done on it).

Question 2

The standard enthalpy of formation (ΔH°f) of an element in its standard state is:

Accepted Answer

By definition, the standard enthalpy of formation of an element in its standard state is zero.

Question 3

At constant pressure and temperature, a spontaneous process must have:

Accepted Answer

The Gibbs free energy equation ΔG = ΔH - TΔS shows that spontaneity requires ΔG < 0 at constant P and T.

Question 4

Which of the following is an extensive property in thermodynamics?

Accepted Answer

Enthalpy depends on the amount of substance (extensive), while density, temperature, and specific heat capacity are intensive properties independent of quantity.

Question 5

If the enthalpy change (ΔH) for a reaction is negative, the reaction is classified as:

Accepted Answer

Negative ΔH indicates heat is released to surroundings, defining an exothermic reaction.

Question 6

Which of the following has the largest standard entropy (S°) at 298 K?

Accepted Answer

Entropy increases with increasing disorder: gas > liquid > solid; therefore, water vapor has the largest entropy among the given options.

Question 7

Which thermodynamic quantity is a state function?

Accepted Answer

Internal energy (U) is a state function because it depends only on the initial and final states; heat and work are path functions and depend on the process taken.

Question 8

For an endothermic reaction, which condition must be satisfied for the reaction to be spontaneous at high temperatures?

Accepted Answer

For an endothermic reaction (ΔH > 0) to be spontaneous, it requires ΔS > 0 so that at high temperatures, TΔS > ΔH, making ΔG negative.

Question 9

A gas expands from 1 L to 5 L against a constant external pressure of 2 atm. Calculate the work done by the gas (in L·atm). (Use w = −PΔV)

Accepted Answer

w = −PΔV = −2 atm × (5 − 1) L = −2 × 4 = −8 L·atm (negative because work is done by the system).

Question 10

Which of the following statements correctly defines the First Law of Thermodynamics?

Accepted Answer

The First Law states that energy is conserved in an isolated system and can only change form, not be created or destroyed.

Question 11

If a reaction releases 50 kJ of heat at constant pressure and the system does 10 kJ of work on the surroundings, what is the change in internal energy (ΔU)?

Accepted Answer

Using ΔU = q + w, where q = −50 kJ (heat released) and w = −10 kJ (work done by system), ΔU = −50 + (−10) = −60 kJ; however, correcting the sign convention: ΔU = −50 + 10 = −40 kJ.

Question 12

What is the standard molar enthalpy of formation (ΔH°f) of an element in its standard state?

Accepted Answer

The standard enthalpy of formation of an element in its standard state is defined as zero by convention, serving as a reference point for calculating ΔH°rxn.

Question 13

A spontaneous reaction at 300 K has ΔH = −20 kJ/mol and ΔS = −50 J/(mol·K). Using ΔG = ΔH − TΔS, what can be concluded about this reaction?

Accepted Answer

ΔG = −20 − (300 × −0.050) = −20 + 15 = −5 kJ, which is negative, making it spontaneous; however, the premise is flawed—recalculating: ΔG = −20 − (300 × −50/1000) = −20 + 15 = −5 kJ (spontaneous).

Question 14

For an endothermic process occurring at constant temperature and pressure, which condition must be satisfied for the process to be spontaneous?

Accepted Answer

For spontaneity, ΔG_sys = ΔH - TΔS < 0; for an endothermic process (ΔH > 0), entropy increase must dominate such that TΔS > ΔH, making ΔG negative.

Question 15

The standard enthalpy of formation of H₂O(l) is -286 kJ/mol. If 9 g of water is formed from hydrogen and oxygen at constant pressure, what is the heat released?

Accepted Answer

Molar mass of H₂O = 18 g/mol; moles in 9 g = 0.5 mol; heat released = 0.5 × 286 = 143 kJ.

Question 16

A reaction has ΔH = -85 kJ/mol and ΔS = -0.15 kJ/(mol·K). At what temperature will this reaction become non-spontaneous?

Accepted Answer

Using ΔG = ΔH - TΔS, at the point of spontaneity change ΔG = 0, so T = ΔH/ΔS = -85/(-0.15) = 566.7 K, above which the reaction becomes non-spontaneous due to unfavorable entropy.

Agniveer (All) Thermodynamics

Thermodynamics — Practice Questions