# Benzene burns according to the following balanced equation: C6H6(l)+15/2O2(g) — 6CO2(g)+3H2O(g)?

0

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

Also Check This  how do you say look over here in japanese?