Which of the following reactions have a positive ÄSrxn? Check all that apply.
A(g) + B(g)--> C(g)
2A(g) + B(g)--> 4C(g)
A(s) + B(g)--> 2C(g)
2A(g) + 2B(g)--> 3C(g)
ok, the postive what?
I think that's dSrxn.
Look at the mols gas on each side. Fewer mols = -dS. More mols = + dS
That's more to fewer mols = -dS
Fewer to more mols = + dS
Would it be the second one that is positive?
To determine which of the reactions have a positive ΔSrxn (change in entropy), we need to consider the number of gas molecules on both sides of the reaction equation. Entropy generally increases when the number of gas molecules increases.
Let's analyze each reaction:
A(g) + B(g) --> C(g)
- This reaction involves one molecule of A and one molecule of B combining to form one molecule of C. There is no net change in the number of gas molecules. Therefore, the ΔSrxn is likely to be close to zero or slightly negative.
2A(g) + B(g) --> 4C(g)
- In this reaction, two molecules of A and one molecule of B combine to form four molecules of C. The number of gas molecules increases from three to four. This increase in gas molecules indicates a positive ΔSrxn.
A(s) + B(g) --> 2C(g)
- This reaction involves a solid A reacting with a gas B to form two molecules of C. Since the solid A does not contribute to the total entropy, we only consider the change in gas molecules. The number of gas molecules increases from one to two, indicating a positive ΔSrxn.
2A(g) + 2B(g) --> 3C(g)
- In this reaction, two molecules of A and two molecules of B combine to form three molecules of C. The total number of gas molecules decreases from four to three. This decrease in gas molecules suggests a negative ΔSrxn.
Based on our analysis, the reactions 2A(g) + B(g) --> 4C(g) and A(s) + B(g) --> 2C(g) have a positive ΔSrxn.