Calculate the enthalpy change for the reaction: 2 h2o2(l) → 2 h2o(l) + o2(g) using enthalpies of formation: δh∘f(h2o2)=−187.8 kj/mol δh∘f(h2o)=−295.8 kj/mol

Ans : 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 ΔH changes for the reaction H​2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ becomes H2O(l) + 1/2 O2(g) → H2O2(l); ΔH = +98.2 kJ Second, multiply this reaction by 2. When multiplying a reaction by a constant, the ΔH is multiplied by the same constant. 2 H2O(l) + O2(g) → 2 H2O2(l); ΔH = +196.4 kJ ANSWER: ΔH = +196.4 kJ for the reaction: 2 H2O(l) + O2(g) → 2 H2O2(l)

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