Consider the following reaction.
2 CO (g) + O2 (g) −→ 2 CO2 (g)
What is most likely true about the entropy change for this reaction?
Delta Srxn < 0 or Delta Srxn > 0.
Why is it Delta Srxn < 0 ?
You have increased order, making three volumes two. Entropy decreases
On the left you have 3 mols of gas. ON the right you have only 2 mols gas. There is more randomness (more random possibilities) in 3 mols than 2 mols.
thank you
To determine the entropy change (ΔSrxn) for a reaction, you need to consider the change in the number of moles of gas molecules. In this reaction, two molecules of CO(g) combine with one molecule of O2(g) to form two molecules of CO2(g).
Initially, you have three moles of gas (2 moles of CO and 1 mole of O2). After the reaction, you have two moles of gas (2 moles of CO2).
Δn = (moles of gas products) - (moles of gas reactants) = 2 - 3 = -1
Since Δn is negative, this means that there is a decrease in the number of moles of gas during the reaction.
According to the second law of thermodynamics, the entropy change for a reaction is directly related to the change in the number of moles of gas. When the number of gas molecules decreases, the entropy decreases.
Therefore, for the given reaction, ΔSrxn < 0.