Consider the following reaction.
2 CO (g) + O2 (g) −→ 2 CO2 (g)
What is most likely true about the entropy change for this reaction?
Why would the delta s of the rxn be less than zero?
thanks.
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The entropy change (ΔS) for a reaction can be determined by looking at the changes in the number of moles of gas molecules between the reactants and products.
In the given reaction, 2 moles of CO (carbon monoxide) react with 1 mole of O2 (oxygen) to produce 2 moles of CO2 (carbon dioxide). The reaction involves the conversion of three gas molecules into two gas molecules.
When the number of gas molecules decreases during a reaction, the entropy generally decreases (ΔS < 0), assuming all other factors are held constant. This is because gases have higher entropy compared to liquids or solids due to their greater molecular freedom and disorderliness.
Therefore, for the given reaction:
2 CO (g) + O2 (g) → 2 CO2 (g)
The reaction involves a decrease in the number of gas molecules (from 3 to 2), indicating a decrease in entropy. Hence, it is likely that the entropy change (ΔS) for this reaction is less than zero.
It's important to note that the entropy change (ΔS) can also be influenced by other factors such as temperature and changes in phase (liquid to gas, solid to liquid, etc.). To determine the exact value of ΔS, you would need to consider these additional factors and use the specific entropy values for the substances involved in the reaction.