Use Hess\'s law and the standard molar heats of formation from the Appendix in y

Question

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 C_6H_6 (l)) for the reaction: 2 C_6H_6 (l) + 15 O_2 (g) rightarrow 12 CO_2 (g) + 6 H_2O(l) Show the individual chemical for the formation of each compound, written in the appropriate direction, multiplied by the needed), showing the enthalpy change of each individual reaction with the correct sign and numerical value.

Explanation / Answer

Enthalpy of formation of Benzene (l):

1. 6C(s) + 3H2(g) --------> C6H6 (l) H = 48.7 kJ/mol

Enthalpy of formation of CO2:

2. C(s) + O2(g) ---------> CO2 (g) H = -393.5 kJ/mol

Enthalpy of formation of H2O(l)

3. H2(g) + 1/2 O2(g) ---------> H2O(l) H = -286 kJ/mol

Combustion of benzene:

4. 2C6H6 + 15O2 ---------> 12CO2 + 6H2O

As seen from equation 4, to use Hess's law, we need to multiply equations 1,2 and 3 to obtain appropriate coefficients of C6H6, CO2 and H2O so that these equations can be used to calculate enthalpy of combustion of benzene using Hess's law.

Multiply equation 1 by 2 (numerical value)

5. 12C(s) + 6H2 (g) --------> 2C6H6 (l) H = 97.4 kJ

Multiply equation 2 by 12 (numerical value)

6. 12 C(s) + 12O2(g) -------> 12 CO2 (g) H = -393.5 x 12 = -4722 kJ

Multiply equation 3 by 6 (numerical value)

7. 6H2(g) + 3O2(g) ------> 6H2O (l) H = -286 x 6 = -1716 kJ

To obtain equation 4 from equation 5, equation 6 and equation 7:

Reversing the equation 5 to obtain Benzene on the reactant side so,

8. 2C6H6 (l) --------> 12C(s) + 6H2 (g) H = - 97.4 kJ

Now adding equation 8,equation 6 and equation 7 we get

2C6H6 (l) + 12 C(s) + 12O2(g) + 6H2(g) + 3O2(g) --------> 12C(s) + 6H2 (g) + 12 CO2 (g) + 6H2O (l) H = (-1716 kJ) + (-4722 kJ) + (- 97.4 kJ) = - 6535.4 kJ

The net equation becomes:

2C6H6 (l) + 12O2(g) ---------> 12 CO2 (g) +  6H2O (l) H = -6535.4 kJ

This value obtained is for moles of Benzene,so for 1 mole of benzene the enthalpy of combustion = -6535.4/2 = -3267.7 kJ/mol of Benzene

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