The standard enthalpy of reaction, $\Delta$H°_rxn, for the reaction,CaO(s)+ SO₃(g)$\rarr$ CaSO₄(s)is -401.5 kJ mol^-1. Determine the value of $\Delta$E°_rxn for this reaction.

Question

Quizplus · Accepted Answer

The change in internal energy, $\Delta$E°_rxn, can be calculated using the relation $\Delta$E = $\Delta$H - \Delta(nRT), where $\Delta$n is the change in moles of gas. Here, $\Delta$n = 0 - 1 = -1. Assuming standard conditions (298 K), $\Delta$E = -401.5 kJ - (-1 mol)(8.314 J/mol·K)(298 K) = -401.5 kJ + 2.48 kJ = -399.0 kJ.

The Standard Enthalpy of Reaction Δ\DeltaΔ H°rxn, for the Reaction,CaO(s)+ SO3(g) →\rarr→

The Standard Enthalpy of Reaction $\Delta$ H°_rxn, for the Reaction,CaO(s)+ SO₃(g) $\rarr$