Postlaboratory Assignment Use Hess\'s law and the standard molar heats of format

Question

Postlaboratory Assignment Use Hess's law and the standard molar heats of formation from the Appendix in your lecture textbook to calculate the standard molar enthalpy of combustion for benzene. That is, calculate delta H degree (in kJ per mole of C6H6(l) for the reaction: Show the individual chemical equation for the formation of each compound. written in the appropriate direction, multiplied by the correct coefficient (where needed), showing the enthalpy change of each individual reaction with the correct sign and numerical value.

Explanation / Answer

Hess Law states that we can arrange and rearrange enthalpy of formations, since this is a state funciton property

C + O2 --> CO2 H = 393.509 kJ/mol

H2 + 1/2O2 --> H2O H = 285.8 kJ/mol

6C(s) + 3H2(g) --> C6H6(l) H = 48.95 kJ/mol

NOTE that for O2 there is no enthalpy of formaiton since it is in its elemental state

We want:

2C6H6(l) + 15O2 --> 12 CO2(g) + 6 H2O(l)

Therefore:

C + O2 --> CO2 H = 393.509 kJ/mol

H2 + 1/2O2 --> H2O H = 285.8 kJ/mol

6C(s) + 3H2(g) --> C6H6(l) H = 48.95 kJ/mol

Send (3) to reverse,

C + O2 --> CO2 H = 393.509 kJ/mol

H2 + 1/2O2 --> H2O H = 285.8 kJ/mol

C6H6(l) --> 6C(s) + 3H2(g) H = -48.95 kJ/mol (change sign to (-))

Multiplye Eqn (1) by 12, Eqn (2) by 6 and Eqn (3) by 2

12C + 12O2 --> 12CO2 H = 12*(393.509 kJ/mol) = -4722.12 kJ

6H2 + 3O2 --> 6H2O H = 6*(285.8 kJ/mol) = -1714.8 kJ

2C6H6(l) --> 12C(s) + 6H2(g) H = 2(-48.95 kJ/mol) = -97.9 kJ

Add all equations

12C + 12O2 + 6H2 + 3O2 +2C6H6(l) -->12CO2 + 6H2O + 12C(s) + 6H2(g) H = -4722.12 + -1714.8-97.9 = -6534.8 kJ

Cancel common terms

15O2 +2C6H6(l) -->12CO2 + 6H2O H = -6534.8 kJ

The equation is what we wanted and the HRxn =  -6534.8 kJ

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