Q.For the reaction at 298 K, $2A + B \rightarrow C$, $\Delta H = 400$ kJ mol$^{-1}$ and $\Delta S = 0.2$ kJ K$^{-1}$ mol$^{-1}$. At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range?

Concept & Intuition

Entropy ($S$) is a measure of the number of microscopic arrangements (disorder) available to a system. A positive $\Delta S$ means the final state is more disordered than the initial state. The biggest entropy increases come from:

• Increasing the number of gas molecules (more particles = more ways to arrange them).
• Changing phase from solid → liquid → gas (gas has far more freedom of motion).
• Increasing volume or temperature.

We’ll check each reaction for these factors.

1. Option (A): $\mathrm{H_2O(l) \rightarrow H_2O(s)}$

   Liquid water freezes into ice. In ice, molecules are locked into a crystal lattice, which is more ordered than the random motion in liquid. So disorder decreases.

   $\Delta S < 0$ — not positive.

2. Option (B): $3\mathrm{O_2(g) \rightarrow 2O_3(g)}$

   Both sides are gases, but the number of gas molecules drops from 3 to 2. Fewer molecules means fewer possible positions and velocities — less disorder.

   $\Delta S < 0$ — not positive.

3. Option (C): $\mathrm{H_2O(l) \rightarrow H_2O(g)}$

   Liquid water becomes water vapor. Gas molecules are far apart, move freely, and occupy a much larger volume — a huge increase in disorder.

   $\Delta S > 0$ — positive. …

The key idea here is the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$. A reaction is spontaneous when $\Delta G < 0$. At the exact temperature where it becomes spontaneous, $\Delta G = 0$. Since $\Delta H$ and $\Delta S$ are given as positive, the reaction is non-spontaneous at low temperatures (because the positive $\Delta H$ dominates) and becomes spontaneous at high temperatures (when the $T\Delta S$ term overtakes $\Delta H$). We simply find the temperature where the crossover happens.

1. Write the condition for spontaneity. The reaction becomes spontaneous when $\Delta G \le 0$.

2. Rearrange to solve for $T$.

3. Plug in the given values. $\Delta H = 400\ \text{kJ mol}^{-1}$, $\Delta S = 2\ \text{kJ K}^{-1}\text{mol}^{-1}$.

4. Interpret the result. …

The core concept here is the First Law of Thermodynamics, which is essentially a statement of the conservation of energy for a thermodynamic system. It relates the change in a system's internal energy ($\Delta U$) to the heat ($q$) exchanged with its surroundings and the work ($w$) done on or by the system.

The internal energy ($U$) of a system is the total energy contained within it, including kinetic and potential energies of its molecules. When a system undergoes a process, its internal energy can change due to two main ways:

1. Heat transfer ($q$): Energy can flow into or out of the system as heat.
2. Work done ($w$): Energy can be transferred as work, either by the system doing work on its surroundings or by the surroundings doing work on the system.

The First Law of Thermodynamics states:

Crucially, we need to establish a consistent sign convention for $q$ and $w$:

• Heat ($q$):
  • If heat is absorbed by the system (endothermic process), $q$ is positive ($q > 0$). This increases the system's internal energy.
  • If heat is released by the system (exothermic process), $q$ is negative ($q < 0$). This decreases the system's internal energy.
• Work ($w$):
  • If work is done on the system by the surroundings (e.g., compression), $w$ is positive ($w > 0$). This increases the system's internal energy.
  • If work is done by the system on the surroundings (e.g., expansion), $w$ is negative ($w < 0$). This decreases the system's internal energy.

Now, let's apply this understanding to the given problem.

1. Identify the heat transfer ($q$): …

We are given standard enthalpies of formation ($\Delta_f H^\Theta$) for three compounds and the enthalpy change for a decomposition reaction. The key idea is Hess’s law: the enthalpy change of a reaction is the difference between the sum of formation enthalpies of products and the sum of formation enthalpies of reactants, regardless of the path.

1. Write the reaction clearly

with $\Delta_r H^\Theta = +175$ kJ mol⁻¹.

2. Recall the relationship For any reaction:

This is a direct consequence of enthalpy being a state function.

3. Plug in the given values

   • $\Delta_f H^\Theta(\text{AO}) = -635$ kJ mol⁻¹
   • $\Delta_f H^\Theta(\text{BO}_2) = x$ kJ mol⁻¹
   • $\Delta_f H^\Theta(\text{ABO}_3) = -1210$ kJ mol⁻¹

   So:

4. Simplify the equation $$$$

Concept & Intuition

Entropy ($S$) is a measure of the randomness or disorder of a system. In thermodynamics, spontaneous processes tend to increase total entropy, but we can still ask whether the system’s entropy goes up or down. The key is to compare the states of matter and the number of particles:

• Solids are more ordered than liquids, which are more ordered than gases.
• More gas molecules mean more disorder (higher entropy).
• Breaking bonds (e.g., diatomic molecule → atoms) increases disorder.

Thus, to find where entropy decreases, look for a process that produces a more ordered state — fewer gas molecules, or a phase change from liquid to solid.

Step-by-step reasoning

1. Process (A): A liquid crystallises into a solid

   • In a liquid, molecules move freely; in a solid, they are locked into a lattice.
   • The solid is far more ordered → entropy decreases.
   • So (A) is a candidate.

2. Process (B): $2\text{NaHCO}_3 (s) \rightarrow \text{Na}_2\text{CO}_3 (s) + \text{CO}_2 (g) + \text{H}_2\text{O} (g)$

   • Reactants: 2 moles of solid. Products: 1 mole of solid + 2 moles of gas.
   • Gases have much higher entropy than solids. The number of gas particles increases from 0 to 2 moles.
   • Overall, disorder increases → entropy increases.
   • So (B) does not show a decrease.

3. Process (C): $\text{H}_2 (g) \rightarrow 2\text{H} (g)$

   • One mole of diatomic gas becomes two moles of monatomic gas. …

The key idea here is the Arrhenius equation:

$k = A e^{-E_a/(RT)}$.

When a catalyst is added, the pre-exponential factor $A$ may change slightly, but the dominant effect is a reduction in the activation energy $E_a$. At the same temperature, the ratio of rate constants depends only on the difference in activation energies.

We are told the rate constant becomes 4 times larger after adding the catalyst. That means:

Since temperature is constant ($27^\circ \mathrm{C} = 300\,\mathrm{K}$), we can write:

Let $\Delta E_a = E_{a,c} - E_{a,u}$ (the change in activation energy). Then:

Taking natural logs on both sides:

We are given $\ln(1/4) = -1.386$, so $\ln 4 = 1.386$. Thus:

Now plug in $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ and $T = 300\ \mathrm{K}$:

First compute $8.314 \times 300 = 2494.2$. Then:

$\Delta E_a = -1.386 \times 2494.2 \approx -3456\ \mathrm{J/mol}$ …