Predict the sign of Delta S in the system for each of the following reactions.
2H3O^+(aq) + CO3^2-(aq) ---> CO2(g) + 3H2O(l)
-,+, or 0?
Delta Srxn = sum So products - sum So reactants.
But the tables in my text (a freshman text) don't have S listed. If your does, it just a simple matter of substituting the numbers and running the above.
To predict the sign of ΔS (the change in entropy) in the system for a given reaction, we need to consider the nature of the reactants and products.
In general, the factors that determine the sign of ΔS include the physical states of the substances involved (solid, liquid, gas, aqueous), the number of molecules or moles of substances, and any changes in the molecular complexity or arrangement.
Let's break down the given reaction:
2H3O+(aq) + CO3^2-(aq) ---> CO2(g) + 3H2O(l)
The reactants consist of two aqueous (aq) solutions, which contain ions, and the products are carbon dioxide (CO2) in its gaseous (g) state and liquid water (H2O) molecules.
The key points to consider are:
1. Aqueous solutions typically have higher entropy compared to solids or liquids. Therefore, the dissolution of ions in water usually leads to an increase in entropy.
2. Gaseous substances tend to have higher entropy than liquids or solids because their molecules are more dispersed and have greater freedom of motion.
Based on these considerations, we can predict the sign of ΔS for each component of the reaction:
2H3O+(aq) + CO3^2-(aq) ---> CO2(g) + 3H2O(l)
- The formation of CO2(g) from the ions in solution involves a decrease in the number of particles and a decrease in molecular complexity (the formation of a gas from ions). Therefore, the ΔS for CO2(g) will be negative (-).
- The formation of 3H2O(l) involves a decrease in the number of particles (from 5 to 3) and a decrease in molecular complexity (the formation of a liquid from ions). Therefore, the ΔS for 3H2O(l) will also be negative (-).
In summary, the predicted sign of ΔS for the given reaction is negative (-) for both CO2(g) and 3H2O(l).