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Question: If \(\Delta H_{f}^{0}\) for \(H_{2}O_{2}\) and \(H_{2}O\) are – 188kJ/mole and \(- 286kJ/mole.\) Wha...

If ΔHf0\Delta H_{f}^{0} for H2O2H_{2}O_{2} and H2OH_{2}O are – 188kJ/mole and −286kJ/mole.- 286kJ/mole. What will be the enthalpy change of the reaction 2H2O2(l)→2H2O(l)+O2(g)2H_{2}O_{2}(l) \rightarrow 2H_{2}O(l) + O_{2}(g)

A

−196kJ/mole- 196kJ/mole

B

146kJ/mole146kJ/mole

C

−494kJ/mole- 494kJ/mole

D

−98kJ/mole- 98kJ/mole

Answer

−196kJ/mole- 196kJ/mole

Explanation

Solution

H2+O2→H2O2H_{2} + O_{2} \rightarrow H_{2}O_{2}; ΔHf0=−188kJ/mole\Delta H_{f}^{0} = - 188kJ/mole

H2+12O2→H2O\mathbf{H}_{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{O}_{\mathbf{2}}\mathbf{\rightarrow}\mathbf{H}_{\mathbf{2}}\mathbf{O} ΔHf0=−286kJ/mole\mathbf{\Delta}\mathbf{H}_{\mathbf{f}}^{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{286kJ/mole}

ΔH=ΔH0(product)−ΔH0(Reactants)\mathbf{\Delta H = \Delta}{\mathbf{H}^{\mathbf{0}}}_{\mathbf{(}\text{product}\mathbf{)}}\mathbf{-}\mathbf{\Delta}{\mathbf{H}^{\mathbf{0}}}_{\mathbf{(}\text{Reactants}\mathbf{)}}

=(2×−286)−(2×−188)=−572+376=−196\mathbf{= (2 \times - 286) - (2 \times - 188)}\mathbf{=}\mathbf{-}\mathbf{572 + 376 =}\mathbf{-}\mathbf{196}