Answer:
+ 636 KJ
Explanation:
We want to arrive to the equation
C6H12O6(s) ---------> 6 H2CO(g) ΔH ° rxn = ?
by manipulating algebraically the first four given equations.
We notice the first one has our product H2CO(g) as a reactant. This indicates we must take the inverse of that equation. Also we need 6 mol of H2CO(g), thus it also needs to be multiplied by 6
6 CO2(g) + 6 H2O(g) ------->6H2CO(g) + 6O2(g) ΔH °comb = + 572.9 KJ/x 6
Now we want C6H12O6(s) as a reactant and it is a product in the second one, therefore lets reverse it
C6H12O6(s) -------> 6 C(s) + 6 H2(g) + 3 O2(g) ΔH ° f = + 1274.4 KJ/mol
Now if take the third equation and multiply it by six we will cancel the C(s) with the above equation
6 C(s) + 6O2(g) ---------> 6 CO2(g) ΔH ° f = - 393.5 KJ/mol x 6
Finally by multiplying the last equation by 6 and adding all the equations we will arrive at our desired one
6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔH ° f = - 285.8 KJ/mol x 6
then lets add them to get ΔH ° rxn:
6 CO2(g) + 6 H2O(g) ------->6H2CO(g) + 6O2(g) ΔH °comb = + 3437.4 KJ
+ C6H12O6(s) -------> 6 C(s) + 6 H2(g) + 3 O2(g) ΔH ° f = + 1274.4 KJ
+ 6C(s) + 6O2(g) ---------> 6 CO2(g) ΔH ° f = - 2361.0 KJ
+6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔHº f = - 1714.8 KJ
<u> </u>
C6H12O6(s) ---------> 6 H2CO(g)
ΔH ° rxn = 3437.4 + 1274.4 - 2361.0 - 1714.8 = 636 KJ
