Chemical Thermodynamics — Practice Questions with Answers

The first law of thermodynamics is a statement of energy conservation, stating that energy can neither be created nor destroyed, only converted from one form to another.

$\Delta U = q + w$, where $\Delta U$ is the change in internal energy, $q$ is heat absorbed by the system, and $w$ is work done on the system.

Extensive properties depend on the amount (quantity) of matter present in the system — examples include mass, volume, and total internal energy.

Intensive properties do not depend on the amount of matter present — examples include temperature, pressure, density, and molar concentration.

Enthalpy of combustion is the enthalpy change when one mole of a substance is completely burned (combusted) in excess oxygen under standard conditions.

Enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their most stable standard states, under standard conditions.

Hess's law states that the total enthalpy change for a chemical reaction is the same, whether the reaction occurs in a single step or through multiple intermediate steps — it depends only on the initial and final states, not the specific pathway taken.

Practical usefulness: This allows chemists to calculate the enthalpy change for reactions that are difficult, dangerous, or impossible to measure directly, by instead combining the known enthalpy changes of other, more easily measurable related reactions (using appropriate addition/subtraction of these reaction equations and their enthalpies).

Entropy reflects the degree of molecular disorder/randomness — in a gas, molecules move freely and randomly throughout a much larger volume, with far greater positional and motional freedom, compared to the more constrained, ordered arrangement of molecules in a liquid (some freedom of movement) or especially a solid (fixed lattice positions, minimal freedom).

This progressively greater molecular freedom and randomness — solid < liquid < gas — corresponds directly to progressively greater entropy in that same order, which is precisely why gases generally have significantly higher entropy than the liquid or solid state of the same substance.

$\Delta G < 0$ (negative): The reaction is spontaneous (thermodynamically favourable) in the forward direction.

$\Delta G > 0$ (positive): The reaction is non-spontaneous in the forward direction (spontaneous in reverse).

$\Delta G = 0$: The system is at equilibrium, with no net driving force in either direction.

State function: A property whose value depends only on the current state of the system, not on the path taken to reach that state. Examples: internal energy ($U$), enthalpy ($H$), entropy ($S$).

Path function: A property whose value depends on the specific path/process taken between two states, not just the initial and final states. Examples: heat ($q$) and work ($w$).

Melting of ice involves a positive $\Delta H$ (heat must be absorbed to break the ordered hydrogen-bonded ice lattice) and a positive $\Delta S$ (liquid water is more disordered/random than solid ice).

Using $\Delta G = \Delta H - T\Delta S$: at temperatures above 0°C (273 K), the $T\Delta S$ term becomes large enough to exceed $\Delta H$, making $\Delta G$ negative (spontaneous melting). Below 0°C, $T\Delta S$ is too small to overcome $\Delta H$, making $\Delta G$ positive (melting is non-spontaneous; freezing is instead favoured) — exactly at 0°C, $\Delta G=0$, representing the equilibrium melting/freezing point.

The "system" is the specific part of the universe being studied; the "surroundings" is everything else outside the system.

Open system: Can exchange both matter and energy with surroundings. Closed system: Can exchange energy but not matter with surroundings. Isolated system: Can exchange neither matter nor energy with surroundings.

By convention, standard enthalpy of formation is defined as zero specifically for an element in its most thermodynamically stable form under standard conditions.

Graphite is the more thermodynamically stable allotrope (form) of carbon under standard conditions compared to diamond — this is precisely why graphite's standard enthalpy of formation is defined as zero, while diamond, being a different (less stable) allotropic form, has a small but non-zero standard enthalpy of formation, reflecting the enthalpy difference between these two distinct forms of elemental carbon.

Convert $\Delta S$ to kJ: $\Delta S = -100\text{ J/K·mol} = -0.1\text{ kJ/K·mol}$

$\Delta G = \Delta H - T\Delta S = -50 - (300)(-0.1)$

$\Delta G = -50 + 30 = \mathbf{-20\text{ kJ/mol}}$

Since $\Delta G$ is negative, the reaction is spontaneous at 300 K.

Target reaction: $C(s)+\dfrac{1}{2}O_2(g)\rightarrow CO(g)$

Using Hess's law: Target = Reaction 1 − Reaction 2

$\Delta H_{target} = \Delta H_1 - \Delta H_2 = -393.5 - (-283) = -393.5+283$

$\Delta H_{target} = \mathbf{-110.5\text{ kJ/mol}}$

$\Delta U = q + w$

With $q=+200\text{ J}$ (heat absorbed) and $w=+50\text{ J}$ (work done ON the system, positive convention):

$\Delta U = 200 + 50 = \mathbf{250\text{ J}}$

The reaction becomes spontaneous when $\Delta G < 0$, i.e., when $\Delta H < T\Delta S$. The boundary (spontaneity threshold) occurs at $\Delta G=0$: $T = \dfrac{\Delta H}{\Delta S}$

Convert to consistent units: $\Delta H = 40000\text{ J/mol}$, $\Delta S=100\text{ J/K·mol}$

$T = \dfrac{40000}{100} = \mathbf{400\text{ K}}$

The reaction becomes spontaneous above 400 K, since positive $\Delta S$ means the $-T\Delta S$ term becomes increasingly negative (favouring spontaneity) as temperature rises.

Since $H_2$ and $O_2$ are elements in their standard states, their enthalpy of formation is zero.

$\Delta H_{reaction} = \Sigma\Delta H_f(\text{products}) - \Sigma\Delta H_f(\text{reactants})$

$\Delta H_{reaction} = 2(-285.8) - [2(0)+1(0)]$

$\Delta H_{reaction} = \mathbf{-571.6\text{ kJ/mol}}$

Work done by the gas $= P\Delta V = 2\times(10-5) = 2\times5 = 10\text{ L·atm}$

Converting to joules: $W = 10\times101.3$

$W = \mathbf{1013\text{ J}}$

(Note: this represents work done BY the gas; by the sign convention where $w$ in $\Delta U=q+w$ represents work done ON the system, this would be entered as $-1013$ J.)

For isothermal reversible expansion of an ideal gas: $\Delta S = nR\ln\left(\dfrac{V_2}{V_1}\right)$

$\Delta S = 1\times8.314\times\ln\left(\dfrac{20}{10}\right) = 8.314\times\ln(2)$

$\ln(2) \approx 0.693$

$\Delta S = 8.314\times0.693 \approx \mathbf{5.76\text{ J/K}}$

$\Delta H_{combustion} = \Sigma\Delta H_f(\text{products}) - \Sigma\Delta H_f(\text{reactants})$

$\Delta H_{combustion} = [2(-393.5)+2(-285.8)] - [1(52.3)+3(0)]$

$\Delta H_{combustion} = [-787-571.6] - [52.3]$

$\Delta H_{combustion} = -1358.6-52.3 = \mathbf{-1410.9\text{ kJ/mol}}$

Convert $\Delta S$: $-198.3\text{ J/K·mol} = -0.1983\text{ kJ/K·mol}$

$\Delta G = \Delta H-T\Delta S = -92.4-(298)(-0.1983)$

$\Delta G = -92.4+59.1 = \mathbf{-33.3\text{ kJ/mol}}$

Since $\Delta G$ is negative, the reaction is spontaneous at 298 K, despite the unfavourable (negative) entropy change, because the strongly favourable (negative) enthalpy change dominates at this temperature.

From $\Delta G = \Delta H - T\Delta S$, rearranging: $\Delta S = \dfrac{\Delta H - \Delta G}{T}$

$\Delta S = \dfrac{-30-(-12)}{400} = \dfrac{-30+12}{400} = \dfrac{-18}{400}$

$\Delta S = -0.045\text{ kJ/K·mol} = \mathbf{-45\text{ J/K·mol}}$