Wednesday, March 11, 2020
Enthalpy Change for a Specific Amount of Reactant
Enthalpy Change for a Specific Amount of Reactant Here is how to determine the change in enthalpy of a chemical reaction with a given amount of reactant. You may wish to review the laws of thermochemistry and endothermic and exothermic reactions​ before you begin. Problem: For the decomposition of hydrogen peroxide, it is known that:H2O2(l) → H2O(l) 1/2 O2(g); ÃŽâ€H -98.2 kJUsing this information, determine ÃŽâ€H for the reaction:2 H2O(l) O2(g) → 2 H2O2(l) Solution: When looking at the second equation, we see it is double the first reaction and in the opposite direction. First, change the direction of the first equation. When the direction of the reaction is changed, the sign on ÃŽâ€H changes for the reaction H 2O2(l) → H2O(l) 1/2 O2(g); ÃŽâ€H -98.2 kJ becomes H2O(l) 1/2 O2(g) → H2O2(l); ÃŽâ€H 98.2 kJ Second, multiply this reaction by 2. When multiplying a reaction by a constant, the ÃŽâ€H is multiplied by the same constant.2 H2O(l) O2(g) → 2 H2O2(l); ÃŽâ€H 196.4 kJ Answer: ÃŽâ€H 196.4 kJ for the reaction: 2 H2O(l) O2(g) → 2 H2O2(l)
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