Use Hess's Law and the following information to calculate the change in enthalpy for the reaction 2C + H2 -> C2H2?

C2H2 + 5/2 O2 -> 2CO2 + H2O Entropy = -1299.6 kJ
C + O2 -> CO2 Entropy = -393.5 kJ
H2 + 1/2 O2 -> H2O Entropy = -285.8 kJ

1 Answer
anor277
Oct 15, 2016

2C(s) +H_2(g) rarr HC-=CH(g)
DeltaH^@""_"reaction"=227*kJ*mol^-1

Explanation:

You have listed entropy changes for the reaction. In fact, these are values for DeltaH^@""_"reaction"

HC-=CH + 5/2O_2 rarr 2CO_2 + H_2O ;DeltaH^@""_"reaction"=-1299.6*kJ*mol^-1 (i)

C + O_2 rarr CO_2 ;DeltaH^@""_"reaction"=-393.5*kJ*mol^-1 (ii)

H_2 + 1/2O_2 rarr H_2O ;DeltaH^@""_"reaction"=-285.6*kJ*mol^-1 (iii)

We want DeltaH^@ for:

2C+H_2 rarr HC-=CH ;DeltaH^@=??

If I do the sum of the given reactions in this manner, 2xx(ii)-(i)+(iii), I get the following:

2C + cancel(2O_2) +cancel(2CO_2) + cancel(H_2O) +H_2 + cancel(1/2O_2)rarr cancel(2CO_2)+HC-=CH +cancel(H_2O)+ cancel(5/2O_2)

Which after cancelling gives:

2C(s) +H_2(g) rarr HC-=CH(g) DeltaH^@""_"reaction"=227*kJ*mol^-1

Note here that DeltaH^@""_"reaction"=DeltaH^@""_f" acetylene".
So here we have used the thermochemical equations as linear equations to give us the equation that we sought (and thus necessarily the thermodynamic parameters).

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