Calculation of standard molar enthalpy of formation

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anto007
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Calculation of standard molar enthalpy of formation

Post by anto007 »

Hi everyone, I tried to do this exercise but I don't get the expected result from my text, would you help me understand where my mistake is? Thank you all.

The text of the exercise says: "If you burn at 25 ° C and 1 atm 0.100 dm ^ 3 of ethane, C2H6 (g), measured at 20 ° C and 1.50 atm you get 9.733 kJ of thermal energy. operates under the same conditions, for combustion of 0.284 g of carbon in the allotropic form of graphite 9.304 kJ of heat are obtained, while for combustion of 1,000 dm ^ 3 of hydrogen, measured at 18 ° C and 0.800 atm, 9.571 kJ are obtained. Calculate the molar enthalpy of ethane formation. "

First of all, I note that the heats reported in the text concern combustions conducted at 25 ° C and 1 atm so that the enthalpies that can be deduced are the standard ones, but I still calculate the combustion enthalpies anyway.

For C2H6 n=PV/RT = (1,50 atm* 0.100 L)/(0.08206 L atm mol^-1 K^-1 * 293.15 K)= 0.00623548 mol;
q=-9.733 kJ; ΔH°comb(C2H6)= -9.733 kJ/0.00623548= -1560.9 kJ mol^-1
For C n=0.284g/12.011g mol^-1= 0.023645 mol;
q=-9.304 kJ; ΔH°comb(C)= -9.304 kJ/0.023645 mol=-393.5 kJ mol^-1
For H n=PV/RT= (0.800 atm*1.000 L)/(0.08206 L atm mol^-1 K^-1*291.15 K)=0.0334843 mol
q=-9.571kJ; ΔH°comb(H)= -9.571kJ/0.0334843 mol= -285.8 kJ mol^-1

I apply Hess's law
4C(graphite)+ 4O2 --> 4CO2 ΔH°a= 4ΔH°comb(C)
6H2(g)+ 3O2(g) --> 6H2O (l) ΔH°b= 6ΔH°comb(H)
4CO2(g)+ 6H2O(l) --> 2C2H6(g)+ 7O2 ΔH°c= -2ΔH°com(C2H6)
____________________________________________________

4C(graphite)+ 6H2(g) --> 2C2H6 (g)
ΔH°f(C2H6)= [(-4 * 393.5)+(-6 * 285.8 )+(2 * 1560.9)]kJ mol^-1= -167 kJ mol^-1

but the result of my text is -84 kJ mol ^ -1, where did I go wrong?
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ChenBeier
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Re: Calculation of standard molar enthalpy of formation

Post by ChenBeier »

You calculate for 4C + 6H2 => 2 C2H6 -167 kJ/mol

But you can devide to 2C + 3H2 => C2H6 -83,55kJ/ mol ~ - 84 kJ/ mol
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anto007
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Re: Calculation of standard molar enthalpy of formation

Post by anto007 »

Right, what a fool I was :oops: !! Thanks so much :D
The mind is like a parachute, it only works if it opens. -A. Einstein
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