(a) What do you expect for the sign of \(\Delta S\) in a chemical reaction in which 3 mol of gaseous reactants are converted to 2 mol of gaseous products? (b) For which of the processes in Exercise 19.11 does the entropy of the system increase?

To determine the sign of the entropy change, \(\Delta S\), we can consider the number of moles of gaseous reactants and the number of moles of gaseous products. The entropy change for a chemical reaction can be given as: \[ \Delta S = S_\text{products} - S_\text{reactants} \] In general, the entropy of a gas is directly proportional to its number of moles, so having more moles of gas results in greater randomness and higher entropy. In the given reaction, we have 3 mol of gaseous reactants and 2 mol of gaseous products. Since the number of moles of reactants is higher than the number of moles of products, the randomness of the system is expected to decrease during the reaction. Therefore, the entropy change will be negative, which means: \[ \Delta S < 0 \]

Unfortunately, we don't have Exercise 19.11 available to address this part of the exercise. However, we can still provide general guidance on how to solve this part when Exercise 19.11 is available: 1. Write down the processes given in Exercise 19.11. 2. For each process, check if it increases the number of moles of gases, the randomness of the system, or whether it's a process where the system moves from a more ordered state to a less ordered state. 3. Identify the processes that fulfill any of these conditions, as these are the ones that lead to an increase in the entropy of the system. 4. Write down the selected processes as the answer to part (b) of the exercise.