Question

Complete combustion of hydrocarbons gives CO 2 and H 2 O as the only products. Use the following enthalples of combustion to calculate the enthalpy of reaction (\Delta H r* n ) for C 6 H 6 ()+3H 2 (g) C 6 H 12 (l) Hint Hess law; write the balanced chemical combustion equations for C 6 H 6 (1),H 2 (g) . C 6 H 12 () . C 6 H 2 (s)=-39 19.8 kJmo * l ^ (- 1) - 285, 90kJmo * l ^ (- 1) C 6 H 6 (s)=-3267.80 kJ mol^ -1 H 2 (g)=

Answer 1

Answer: Reaction 1: Reaction 2: Reaction 3: C6H6 (1) + 15/2 02 (g) → 6CO2 (g) + 3H20 (1) H2(g) + 1202 + H2O (1) C6H12 (1) + 902 (g) →6002 (g) + 6H20 (1) C6H6 (1) + 3H2(g) → CH12 (1) AH° = - 3267.80 kJ mol! AH = - 285.90 kJ mol-1 AH = - 3919.80 kJ mol-1 AH° = ? To get the reaction 4, we need to multiply the reaction 2 by 3 and rewrite the reaction 3 in reverse direction Hence, we can write as follows: Reaction 1: C6H6 (1) + 15/2 02 (g) → 6CO2 (g) + 3H20 (1) AH° = - 3267.80 kJ mol-1 Reaction 2: 3H2 (g) + 3/2 02 → 3H20 (1) AH° = -(3x285.90) kJ moll Reaction 3: 6CO2 (g) + 6H20 (1) ► C6H12 (1) +902 (g) AH° = + 3919.80 kJ moll Now adding reaction (1) + (2) + (3) we can obtain reaction (4). According to Hess's Law of constant heat summation the total enthalpy changes will be the sum of enthalpy changes of multiple steps Thus, Reaction 1 +Reaction 2 + Reaction 3 Reaction 1: C6H6 (1) + 15120249) →600248) + 3712047 AH° = -3267.80 kJ mol? Reaction 2: 3H2(g) + 31202 → 3100 AH = - 857.70 kJ mol* Reaction 3: 660249)+612044 → C6H12 (1) +902-tg/ AH° = + 3919.80 kJ mol' Reaction 4; C6H6 (1) + 3H2(g) → C6H12 (1) AH° = (- 3267.80 - 857.70+ 3919.80) kJ mol-1 = -205.70 kJ mol? Therefore, the enthalpy change of reaction 4 is -205.70 kJ mol"