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
Oct 15, 2016

Answer:

#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).