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13 Dec 2019

When one mole of benzene, C6H6 is burned in O2 to produce CO2 and H2O, H = -3300 kJWhat is (Triangle H) for:12 CO2 + 6 H2O ---> 2 C6H6 + 15 O2 answer is in kJ

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Sixta Kovacek
Sixta KovacekLv2
17 Dec 2019
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Related questions

The standard enthalpy of formation of benzene, C6H6, is +49.2 kJ/mol. Calculate the heat of combustion of one mole of C6H6. The heats of formation of CO2(g) and H2O(l) are -394 kJ/mol and -285.8 kJ/mol respectively. 

Using standard enthalpies of formation from Appendix L,calculate ?Ho for the following reaction: 2 C6H6(l) + 15 O2(g) ? 12CO2(g) + 6 H2O(g) b. How many grams of C6H6(l) must be combusted inthe above reaction to produce 2570 kJ of heat energy?

C6H6(l)= 48.95
o2 = O
CO2 = -393.509
H2O = -241.83

8. What is the change in entropy when 5.46 mL of liquid benzene (C6H6, d = 0.879 g/mL) is combusted in the presence of 27.3 L of oxygen gas, measured at 298 K and 1 atm pressure? (R = 0.0821 L · atm/(K · mol))
2 C6H6(l) + 15 O2(g)  12 CO2(g) + 6 H2O(l) ; S° = –437.7 J/K at 298 K
A) –488 J/K B) –46.0 J/K C) –26.9 J/K D) –13.4 J/K E) –515 J/K

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