2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)

a. Determine the enthalpy change for the reaction.

ΔH = (4(-393.5)+ 2(-241.8)) – (2(227) + 5(0))

The enthalpy change is -2511.6.

b. Determine the entropy change for the reaction.

ΔS = (4(213.7)+ 2(188.7)) – (2(200.9) + 5(205))

The entropy change is -194.6.

my teacher told me that my answers were wrong. What did I do wrong?

Thank you for showing your work.

I don't see anything wrong with your set up or the numbers. Always check to see that you used the correct value from the tables. I suspect the problem is one of significant figures. I think the answer is -2512 kJ.
-1.574E3 - 0.4836E3 = -2.058E3 since that last 6 on 483.6 can't be used.
Add that to -454 kJ to obtain -2512 kJ.

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Hmm, it seems like your answers were a bit... off. Let me break it down for you in a funnier way!

a. Enthalpy change: ΔH = (4(-393.5) + 2(-241.8)) - (2(-394) + 5(0))

You actually forgot to change the enthalpy of formation for CO2. It's -394 kJ/mol, not -227 kJ/mol. Once you recalculate it, you'll get the correct enthalpy change.

b. Entropy change: ΔS = (4(213.7) + 2(188.7)) - (2(130.6) + 5(205))

In this case, you mistakenly used the wrong entropy values. The entropy of formation for CO2 is 130.6 J/mol·K, not 200.9 J/mol·K. Fix that, and your answer should be spot on.

Remember, chemistry can be tricky, so double-check your values and rock that test!

To determine what went wrong, let's break down the equation and individual steps in calculating the enthalpy and entropy change.

The given chemical equation is:

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

a. Enthalpy Change (ΔH):

To calculate the enthalpy change, you need to use the standard enthalpies of formation (∆Hf) for each compound involved. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

The standard enthalpies of formation (in kJ/mol) are:

∆Hf(C2H2(g)) = 227
∆Hf(O2(g)) = 0
∆Hf(CO2(g)) = -393.5
∆Hf(H2O(g)) = -241.8

Now, let's calculate the enthalpy change:

∆H = (4 * ∆Hf(CO2(g)) + 2 * ∆Hf(H2O(g))) - (2 * ∆Hf(C2H2(g)) + 5 * ∆Hf(O2(g)))
∆H = (4 * -393.5 + 2 * -241.8) - (2 * 227 + 5 * 0)
∆H = (-1574 - 483.6) - (454 + 0)
∆H = -2057.6 - 454
∆H = -2511.6 kJ/mol

Your calculation for the enthalpy change is correct, so it seems like your teacher may have made a mistake or there might be a misunderstanding.

b. Entropy Change (ΔS):

To calculate the entropy change, you need to use the standard entropies (∆S) of the compounds involved. The standard entropy is a measure of the disorder or randomness of a substance at a given temperature and pressure.

The standard entropies (in J/(mol·K)) are:

∆S(C2H2(g)) = 200.9
∆S(O2(g)) = 205
∆S(CO2(g)) = 213.7
∆S(H2O(g)) = 188.7

Now, let's calculate the entropy change:

∆S = (4 * ∆S(CO2(g)) + 2 * ∆S(H2O(g))) - (2 * ∆S(C2H2(g)) + 5 * ∆S(O2(g)))
∆S = (4 * 213.7 + 2 * 188.7) - (2 * 200.9 + 5 * 205)
∆S = (854.8 + 377.4) - (401.8 + 1025)
∆S = 1232.2 - 1426.8
∆S = -194.6 J/(mol·K)

Your calculation for the entropy change is also correct.

If your teacher claims that your answers are wrong, it's recommended to consult with them again and clarify the specific errors or misconceptions. It's possible that there might be a misunderstanding or a mistake on their part.

I think the problem is your you shouldn't multiply the number of moles with the the elements number ,,,that's what I think ,,