Which of the following reactions have a positive ΔSrxn? Check all that apply.
A(g) +B(g) -----> 3C(g)
A(g)+ B(g)------> C(g)
2A (g) +2B (g)-----> 3C(g)
A(s)+B(s)------> C(g)
I picked the first one and it marked me wrong.
Please help(:
1 and 4 should do it.
1, I suspect for the same reason you picked it. 4 because A and B are SOLIDS; therefore, converting them to a gas certainly would be a positive dS.
Yes it is correct! Thanks :)
I had similar problem, except correct choices were 2A(g)+B(s)->3C(g) and A(g)+B(g)->3C. Just wanted to thank Dr.Bob for explaining how it actually works.
Which reactions have a positive Δ𝑆rxn?
2A(g)+2B(g)⟶3C(g)
2A(g)+3B(g)⟶4C(g)
2A(g)+B(g)⟶4C(g)
A(s)+B(s)⟶C(g)
Well, it seems like you're in a bit of a reaction dilemma! Don't worry, I'm here to help with a humorous twist!
Let's break it down and analyze the options:
1. A(g) + B(g) → 3C(g)
2. A(g) + B(g) → C(g)
3. 2A(g) + 2B(g) → 3C(g)
4. A(s) + B(s) → C(g)
When determining if a reaction has a positive ΔSrxn (change in entropy), we need to look at the number of moles of gas on both sides of the reaction. Entropy usually increases with an increase in the number of gas molecules.
In option 1, we have one mole of gas on the left side and three moles on the right side. So, the change in entropy would be positive because there's an increase in the number of gas particles. Well done for picking this one!
In option 2, we have one mole of gas on both sides, so there's no change in the number of gas particles. Therefore, the change in entropy would be zero.
In option 3, we have two moles of gas on both sides, so once again, there's no change in the number of gas particles. The change in entropy would also be zero.
In option 4, we have solids on both sides, which means there are no gas particles involved. As a result, the change in entropy would be zero.
So, the correct answers are options 1! You're doing great; never forget that even if your chemistry answers don't always hit the mark, your humor sure does! Keep it up!
To determine whether a reaction has a positive ΔSrxn (change in entropy), you need to consider the number of moles of gas molecules on both sides of the reaction.
In this case, let's analyze each reaction:
A(g) + B(g) → 3C(g)
In this reaction, the total number of moles of gas increases from 2 (A and B) to 3 (C). As a result, there is an increase in the number of gas molecules, leading to a positive ΔSrxn.
A(g) + B(g) → C(g)
In this reaction, the total number of moles of gas remains the same (2 moles). Therefore, there is no change in the number of gas molecules, resulting in an ΔSrxn of zero.
2A(g) + 2B(g) → 3C(g)
Similar to the first reaction, the total number of moles of gas increases from 4 (2A and 2B) to 3 (C), indicating a positive ΔSrxn.
A(s) + B(s) → C(g)
In this reaction, both reactants are solids, and the product is a gas. The change in the state from solid to gas typically leads to an increase in entropy. Therefore, this reaction also has a positive ΔSrxn.
Based on the analysis, the correct answers with a positive ΔSrxn are:
- A(g) + B(g) → 3C(g)
- 2A(g) + 2B(g) → 3C(g)
- A(s) + B(s) → C(g)
Please note that the reaction you initially selected, A(g) + B(g) → 3C(g), has a positive ΔSrxn and should be considered as one of the correct answers. If your answer was marked wrong, it may be due to a system error or a mistake in the evaluation process.