dHrxn = (n*dH products) - (n*dH reactants)
-3226.7 kJ = [(3*-393.5)+(3*-385.8)] - [(1*dHbenzoic acid) - (0)] and solvle for dHbenzoic acid.
DrBob222 PLEASE HELP
Hf of CO2 = -393.5 kJ/mol
Hf of H2O = -285.8 kJ/mol
The combustion reaction for benzoic acid C6H5CO2H(s) + 15/2 O2(g) →
7 CO2(g) + 3 H2O(ℓ) has ∆H0 = −3226.7 kJ/mole. Use Hess’s Law to calculate ∆H0f for benzoic acid.
1. −698.1 kJ/mol
2. −348.6 kJ/mol
3. −1609 kJ/mol
4. No choice is correct.
5. −237.4 kJ/mol
6. −385.2 kJ/mol
3 answers
385.2
I made several typos. Here is the corrected.
-3226.7 = [7*-393.5)+(3*-285.8)] - [dHbenzoic acid)-0]
I get 385.2 kJ but a NEGATIVE sign. Check that.
-3226.7 = [7*-393.5)+(3*-285.8)] - [dHbenzoic acid)-0]
I get 385.2 kJ but a NEGATIVE sign. Check that.