Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

1. For a process to be spontaneous, \( \Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0 \).

2. Consider the reaction: \(2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(l)}\). When 2 moles of \(H_{2(g)}\) and 1 mole of \(O_{2(g)}\) gas combine to form 2 moles of liquid water, 572 kJ of heat is released, which is received by the surroundings at constant pressure (and 298 K).

3. The entropy change of the surroundings is given by:

\[ \Delta S_{\text{surr}} = \frac{Q_{\text{rev}}}{T} = \frac{572 \times 10^3 \, \text{J}}{298 \, \text{K}} = +1919 \, \text{J K}^{-1} \]

4. The total entropy change is:

\[ \Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \]

\[ = -327 \, \text{J K}^{-1} + 1919 \, \text{J K}^{-1} \]

\[ = +1592 \, \text{J K}^{-1} \]

5. Since \( \Delta S_{\text{total}} > 0 \), the reaction is spontaneous at 25°C.

6. It follows that to decide the spontaneity of reactions, we need to consider the entropy of the system and its surroundings.