Please help, I have no clue where to start with this problem! Thanks in advance !
1.Calculate DeltaHrxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid benzene is 49.1.)
And also I was stuck on this question too.
2. Equation is 6CO2(g) + 6H2O(l) — C6H12O6(s)+6O2(g)
Calculate DeltaHrxn of photosynthesis
3 Answers

For the first reaction you need to know the entalpies of formation of CO2 (395.5 kJ/mol) and H2O(g) (241.8 kJ/mole).
C6H6(l)+15/2O2(g) — 6CO2(g)+3H2O(g)
dHrn = dHfCO2(g) x 6 + dHf H2O(g) x 3 – dHf C6H6(l)
= 2373 kJ – 725.4 kJ + 49.1 kJ = 3049 kL/mol
For the second reaction you need dHf for glucose = 1271 kJ/mol
6CO2(g) + 6H2O(l) — C6H12O6(s)+6O2(g)
Note that here the equation shows H2O(l) not H2O(g) as in the first equation!!
dHrn = dHf C6H12O6(s) – 6 x dHf CO2(g) – 6 x dHf H2O(l)
= 1271 kJ () 6 x 395.5 kJ () 6 x – 285.8 kJ = +2805 kJ/mol

The right answer is 3135.5 kJ

HELLO ANYONE ELSE NOTICE HOW MANY THUMBS DOWN THERE ARE? THAT IS NOT THE CORRECT ANSWER