Asked by Morgan
                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?
            
            
        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?
Answers
                    Answered by
            DrBob222
            
    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.
                    Answered by
            zoey
            
    I think the problem is your you shouldn't multiply the number of moles with the  the elements number ,,,that's what I think ,,
    
                    Answered by
            randy boylan
            
    dsadasdasdsdadadsadadasdaddasdsdsdd
    
                    Answered by
            Anonymous
            
    a photon of light of 450nm, when compared to light of wavelength 300nm has
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