Q.For the reaction $2A(g) + B(g) \rightarrow 2D(g)$, $\Delta U = -10.5$ kJ and $\Delta S = -44.1$ JK$^{-1}$. Calculate $\Delta G$ for the reaction, and predict whether the reaction may occur spontaneously.

Concept and intuition

Spontaneity is governed by the Gibbs free energy change:

A process is spontaneous when $\Delta G < 0$. This depends on both the enthalpy change $\Delta H$ and the entropy change $\Delta S$ of the system, as well as the temperature $T$. The statement in (C) claims that if $T\Delta S$ is positive, the process is always spontaneous at all temperatures — but that ignores the possibility that $\Delta H$ could be large and positive, making $\Delta G > 0$. Let’s check each option carefully.

Step-by-step reasoning

1. Option (A):

   For a spontaneous process, $\Delta G < 0$. Since $\Delta G = \Delta H_{\text{system}} - T\Delta S_{\text{system}}$, the condition $\Delta H_{\text{system}} - T\Delta S_{\text{system}} < 0$ is exactly the criterion. So (A) is correct.

2. Option (B):

   Entropy depends on the amount of substance (extensive) and its value is determined solely by the state of the system (state function). This is a standard thermodynamic fact. So (B) is correct.

3. Option (C):

   The statement says: “A process will always be spontaneous at all temperatures, if $T\Delta S$ is positive.”

   • If $T\Delta S > 0$, then $\Delta S > 0$ (since $T > 0$).
   • But $\Delta G = \Delta H - T\Delta S$. For spontaneity we need $\Delta G < 0$, i.e., $\Delta H < T\Delta S$.
   • If $\Delta H$ is also positive and larger than $T\Delta S$, then $\Delta G > 0$ and the process is non-spontaneous. …

Concept & Intuition

A reaction is spontaneous (under constant pressure and temperature) when the Gibbs free energy change $\Delta G$ is negative. The relation is

Here $\Delta H = 35.5\ \mathrm{kJ\ mol^{-1}}$ (positive) and $\Delta S = 83.6\ \mathrm{J\ K^{-1}\ mol^{-1}}$ (positive).

• At low temperatures, the $T\Delta S$ term is small, so $\Delta G$ is positive → non‑spontaneous.
• At high temperatures, the $T\Delta S$ term dominates, making $\Delta G$ negative → spontaneous. The crossover happens when $\Delta G = 0$, i.e. $T = \Delta H / \Delta S$. Above that temperature, spontaneity kicks in.

Step‑by‑step reasoning

1. Write the condition for spontaneity Spontaneous when $\Delta G < 0$:

Since $\Delta S > 0$, we can divide without flipping the inequality:

2. Convert units so they match

   $\Delta H = 35.5\ \mathrm{kJ\ mol^{-1}} = 35\,500\ \mathrm{J\ mol^{-1}}$

   $\Delta S = 83.6\ \mathrm{J\ K^{-1}\ mol^{-1}}$

   (Both per mole, so the “per mole” cancels.)

3. Compute the threshold temperature

The key is to read the reaction carefully: “A₂ + 3 B₂ + Heat → 2 AB₃”. The “+ Heat” on the left means heat is absorbed — the reaction is endothermic. That immediately tells us ΔH > 0.

Spontaneity at a given temperature is governed by the Gibbs free energy change:

For a process to be spontaneous, we need ΔG < 0.

Now, if ΔH > 0 (endothermic), the term –TΔS must be negative enough to overcome the positive ΔH. That requires ΔS > 0 (entropy increase), because then –TΔS is negative.

Let’s check each option:

1. Option (A): ΔH < 0, ΔS > 0, ΔG > 0

   • ΔH < 0 is exothermic — contradicts the “+ Heat” clue. Also ΔG > 0 means non-spontaneous. So wrong.

2. Option (B): ΔH > 0, ΔS > 0, ΔG < 0

   • ΔH > 0 matches endothermic. ΔS > 0 gives a negative –TΔS term. At a sufficiently high temperature, |TΔS| > |ΔH|, so ΔG < 0. This is exactly the condition for spontaneity. Correct.

3. Option (C): ΔH > 0, ΔS < 0, ΔG < 0 …

The key concept here is Gibbs free energy change and the condition for spontaneity at constant pressure and temperature. A reaction is spontaneous when ΔG° < 0. The relation ΔG° = ΔH° – TΔS° tells us that if ΔH° is positive (endothermic) and ΔS° is positive (increase in disorder), the reaction will become spontaneous above a certain temperature where the entropy term outweighs the enthalpy term.

1. Identify the condition for spontaneity

   For a process at constant pressure and temperature, spontaneity requires ΔG° < 0.

   Given ΔH° = +179.1 kJ mol⁻¹ and ΔS° = +160.2 J K⁻¹ mol⁻¹, both positive, the reaction is non-spontaneous at low temperatures but becomes spontaneous at high temperatures.

2. Set up the inequality

   We want:

Rearranging:

3. Convert units consistently ΔH° is in kJ, ΔS° is in J. Convert ΔH° to J: $$\Delta H^\circ = 179.1 \text{ kJ mol}^{-1} = 179100 \text{ J mol}^{-1} …$$

The problem gives us the standard enthalpy change ($\Delta H^\circ$) and standard entropy change ($\Delta S^\circ$) for a reaction at 298 K, and asks for the standard free energy change ($\Delta G^\circ$). The relationship between these three thermodynamic quantities is given by the Gibbs-Helmholtz equation:

This equation tells us whether a reaction is spontaneous under standard conditions: if $\Delta G^\circ$ is negative, the reaction is spontaneous; if positive, non-spontaneous. Here, we simply need to plug in the numbers — but careful: the units of $\Delta H^\circ$ and $\Delta S^\circ$ are different (kJ vs. J), so we must convert them to the same unit before subtracting.

Let’s work through it step by step.

1. Identify the given values

   • $\Delta H^\circ = +29.3\ \text{kJ/mol}$
   • $\Delta S^\circ = +104.1\ \text{J K}^{-1} \text{mol}^{-1}$
   • Temperature $T = 298\ \text{K}$

2. Convert entropy to kJ units

   Since $\Delta H^\circ$ is in kJ, we convert $\Delta S^\circ$ from J to kJ:

3. Compute $T \Delta S^\circ$

4. Apply the Gibbs free energy formula $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ = 29.3 - 31.0218 = -1.7218\ \text{kJ/mol}$ …

Step 1 — The concept: Hess's law.

Enthalpy $H$ is a state function: its change depends only on the initial and final states, never on the route taken. Two consequences follow, and this question uses both:

1. Enthalpy changes along a multi-step path simply add.
2. Reversing a step flips the sign of its $\Delta H$.

Step 2 — Read the cycle.

The energy diagram links the three species in sequence, $A \xrightarrow{\ \Delta H_1\ } B \xrightarrow{\ \Delta H_2\ } C$, and closes the loop with the unknown step $C \xrightarrow{\ \Delta_r H\ } A$. Because the loop returns to its starting point, the total enthalpy change around it is zero:

Step 3 — Get $A \rightarrow C$ first.

Both steps are endothermic (positive), so $C$ sits $35\,\mathrm{J}$ above $A$ on the energy diagram.

Step 4 — Reverse it. …

1. Set up the variables from the given ratio

Bond enthalpies are in the ratio $A_2 : B_2 : AB = 2 : 1 : 2$. Let

(all positive, since a bond enthalpy is the energy required to break one mole of that bond).

2. Write the formation reaction of AB

The enthalpy of formation always refers to one mole of the compound made from its elements in their standard states:

The $\tfrac12$ coefficients are essential — a very common slip is to use whole moles of $A_2$ and $B_2$.

3. Apply the bond-enthalpy formula

Substituting:

4. Solve …

Step 1 — Sign of $\Delta G$.

Adsorption of a gas on a solid happens on its own; it is a spontaneous process. Hence

Step 2 — Sign of $\Delta S$ (the key physical insight).

A gas molecule roams freely in three dimensions. Once adsorbed it is held on the surface and loses most of that translational freedom — the system becomes more ordered:

Step 3 — Deduce the sign of $\Delta H$ from the Gibbs–Helmholtz relation.

Rearranged: $\Delta H = \Delta G + T\Delta S$. Both terms on the right are negative ($\Delta G<0$, and $T\Delta S<0$ since $T>0$ and $\Delta S<0$). Therefore

$\Delta H < 0 \quad \text{— adsorption is always EXOTHERMIC.}$ …