WA 7 Dearing

8. From the standard enthalpies of formation, calculate ΔH° rxn for the reaction C 6 H 12 ( l ) + 9O 2 ( g ) → 6CO 2 (g) + 6H 2 O( l ) For C 6 H 12 ( l ), ΔH° f = –151.9 kJ/mol (5 points) Substance ∆H° f , kJ/mol C 6 H 12 ( l ) –151.9 O 2 ( g ) 0 H 2 O( l ) –285.8 CO 2 ( g ) –393.5 Reactants, C 6 H 12 ( l ) – (-151.9) + O 2 (g) 9 x (0) = -3,924.11 kJ/mol Products, CO 2 ( g ) - 6 x (-393.5) + H 2 O( l ) - 6 x (-285.8) = -4087.8 Sum of P – sum of R = (-4087.8) – (-151.9), ΔH°rxn = -3,924.11 kJ/mol 9. Determine the amount of heat (in kJ) given off when 1.26 × 10 4 g of ammonia are produced according to the equation (Reference: Chang 6.59) N 2 ( g ) + 3H 2 ( g ) → 2NH 3 ( g ) Δ H °= –92.6 kJ/mol Assume that the reaction takes place under standard conditions at 25 o C. (5 points) (1.26 × 10 4 ) / 17amu = 741mol NH3, 741mol x ( -92.6 kJ/mol / 2 mol) = -3.43 x 10 4 kJ 10. From the following heats of combustion, (Reference: Chang 6.63) CH 3 OH(l) + 3/2O 2 (g) → CO 2 (g) + 2H 2 O( l ) ΔH o rxn = –726.4 kJ/mol C(graphite) + O 2 ( g ) → CO 2 ( g ) ΔH o rxn = –393.5 kJ/mol H 2 ( g ) + ½O 2 ( g ) → H 2 O( l ) ΔH o rxn = –285.8 kJ/mol Calculate the enthalpy of formation of methanol (CH 3 OH) from its elements. C(graphite) + 2H 2 ( g ) + ½O 2 ( g ) → CH 3 OH( l ) (10 points) C(graphite) = (-393.5) H2 = 2(-285.8) = (-571.8) CH 3 OH(l) = CO 2 (g) + 2H 2 O( l ) → CH 3 OH(l) + 3/2O 2 (g) = 726.4 Sum of all reactions = (-393.5) + (-571.8) + 726.4 = -238.9kJ/mol 11. Methanol (CH 3 OH) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol: (5 points) (Reference: Chang 6.87) 2CH 3 OH( l ) + 3O 2 ( g ) → 2CO 2 ( g ) + 4H 2 O( l ) ΔH o rxn = –1452.8 kJ/mol CO 2 2(-393.5) + H 2 O 4(-285.8) = 1930.2 – (-1452.8) = 477.4 kJ/mol 3