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

NetherCraft 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

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