Question

From the enthalpies of reaction 2H2(g)+O2(g)→2H2O(g)ΔH=−483.6kJ3O2(g)→2O3(g)ΔH=+284.6kJ calculate the heat of the reaction 3H2(g)+O3(g)→3H2O(g) Express your answer using four significant figures. I have no idea how to approach this problem.

Answer 1

2H2(g)+O2(g)→2H2O(g)          ΔH₁=−483.6kJ   .....(1)

3O2(g)→2O3(g)                     ΔH₂=+284.6kJ    ......(2)

calculate the heat of the reaction

3H2(g)+O3(g)→3H2O(g)         ΔH₃= ?    ..................(3)

So we have to calcualte enthalpy of reaction (3) from given enthalpies of reaction (1) and (2)

We will combine the two equations so that we can obtain the third equaiton

Equation (3) = 1.5 X (equation (1)) - (0.5 X Equation (2))

Enthalpy ((3) = 1.5 X enthalpy of (1) - 0.5 X enthalpy of (2)

ΔH₃= 1.5 X (−483.6kJ) - (0.5 X +284.6kJ) = -725.4 - 142.3 = -867.7 kJ