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?

2 Answers

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.

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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