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Question
The enthalpy of reaction for the reaction: \[\ce{2H2 (g) + O2(g) -> 2H2O (l)}\] is ∆rHΘ = – 572 kJ mol–1. What will be standard enthalpy of formation of \[\ce{H2O (l)}\]?
Solution
Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given equation is – 572 kJ mol–1.
Now the enthalpy of formation for \[\ce{H2O}\] will be half the enthalpy of the value in the given equation. So, now we can calculate that:
\[\ce{Δ_rH = 1/2 × Δ_yH = 572/2 = - 286 kJ mol^{-1}}\]
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