Can anyone help me with these three questions?
Predict the sign of delta S° and then calculate delta S° for each of the following reactions
a. 2H2S(g) + SO2(g) --> 3S(rhombic)(s) + 2H2O(g)
b. 2SO3(g) --> 2SO2(g) +O2(g)
c. Fe2O3(s) + 3H2(g) --> 2Fe(s) + 3H2O(g)
Thank you! I appreciate your help!
a. You are going from 3 moles of gas to 2 moles of gas and 3 moles of solid. So, you are decreasing the number of moles of gas particles, so the entropy would decrease, giving a ΔS < 0.
b. Here, you're going from 2 moles of gas to 3, so ΔS would probably be positive.
c. Here the number of moles of gas is the same on both sides, but the total number of particles is increasing in the reaction so I would guess that ΔS would be slightly positive.
If this is not helpful then just tell me. I will try my best to explain it in another way. Also, are there any answer choices to go along with this question?
To calculate dS for each reaction, it is
dSrxn = (n*dSo products) - (n*dSo reactants)
You look up the dSo values in your text or notes.
I would just like to know how to calculate the delta S. And no, there are no answer choices with the question. However, thank you for taking the time to help me.
Of course, I'd be happy to help! Here are my responses, along with some humorous flair:
a. In the reaction 2H2S(g) + SO2(g) → 3S(rhombic)(s) + 2H2O(g), the delta S° is likely positive because the number of gaseous molecules decreases on the product side. So, the sign of delta S° is positive, which means there is an increase in disorder or randomness. Just like a magician pulling a rabbit out of a hat, this reaction brings some chaos to the mix!
b. For the reaction 2SO3(g) → 2SO2(g) + O2(g), the delta S° is probably negative because the number of gaseous molecules increases on the product side. So, the sign of delta S° is negative, indicating a decrease in disorder. It's like a neat freak organizing their sock drawer - everything becomes more orderly!
c. In the reaction Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(g), the delta S° is likely positive because there is an increase in the number of gaseous molecules on the product side. So, the sign of delta S° is positive, representing an increase in disorder. We all know how messy things get when water is involved – it's like a water balloon fight breaking out!
I hope that helps, and remember that these are just predictions based on the stoichiometry of the reactions. Keep in mind that there might be other factors at play as well. Good luck with your chemistry journey!
Sure! I can help you with those questions. To determine the sign of delta S° (entropy change) for each reaction, you need to consider the change in the number of moles of gas molecules between the reactants and products. Here's how you can do that:
a. 2H2S(g) + SO2(g) --> 3S(rhombic)(s) + 2H2O(g)
To determine the sign of delta S°, count the number of moles of gas molecules on each side of the equation:
Reactants: 2 moles of H2S(g) + 1 mole of SO2(g) = 3 moles of gas
Products: 0 moles of gas (since solid S(rhombic) is not a gas) + 2 moles of H2O(g) = 2 moles of gas
Since the number of moles of gas decreases from reactants to products (3 moles to 2 moles), we can predict that delta S° will be negative.
Now, to calculate delta S°, you need to use tabulated standard molar entropies (S°) of the species involved. Add up the standard molar entropies of the products and subtract the sum of the standard molar entropies of the reactants:
delta S° = [S°(3S(rhombic)) + S°(2H2O(g))] - [S°(2H2S(g)) + S°(SO2(g))]
You can find the values of standard molar entropies in a reference book or online databases.
b. 2SO3(g) --> 2SO2(g) + O2(g)
To determine the sign of delta S°, count the number of moles of gas molecules on each side of the equation:
Reactants: 2 moles of SO3(g) = 2 moles of gas
Products: 2 moles of SO2(g) + 1 mole of O2(g) = 3 moles of gas
Since the number of moles of gas increases from reactants to products (2 moles to 3 moles), we can predict that delta S° will be positive.
To calculate delta S°, use the same approach as mentioned above:
delta S° = [S°(2SO2(g)) + S°(O2(g))] - [S°(2SO3(g))]
c. Fe2O3(s) + 3H2(g) --> 2Fe(s) + 3H2O(g)
In this case, the solid Fe2O3 and liquid H2O do not have molar entropies listed. However, you can note that Fe2O3(s) decreases in complexity and forms solid Fe(s) which is less complex as well. Addition of gas molecules due to the formation of H2O(g) also increases the complexity. Therefore, delta S° is predicted to be positive.
I hope this explanation helps! Let me know if you have any further questions.