The standard enthalpy of formation of gaseous water is -241.82 kJ.mol -1 at 298 K .Estimate this value at 370 K.The molar heat capacities of hydrogen,Oxygen,And gaseous waterare 28.82,29.36,And 33.58J.k.mol -1

Question

Quizplus · Accepted Answer

We can use the equation: ΔH₂₉₈ = ΔH₃₇₀ + ∫Cp dT where, ΔH₂₉₈ is the standard enthalpy of formation at 298 K ΔH₃₇₀ is the standard enthalpy of formation at 370 K Cp is the molar heat capacity dT is the temperature difference (370 K - 298 K = 72 K) Substituting the given values, we get: F1F1F1S1F1F1F10241.82 kJF1F1F1S1F1F1F10molF1F1F1S1^F1F1F10¹ = ΔH₃₇₀ + ∫Cp dT Solving for ΔH₃₇₀: ΔH₃₇₀ = F1F1F1S1F1F1F10242.54 kJF1F1F1S1F1F1F10molF1F1F1S1^F1F1F10¹ Therefore, the best choice is C.

The Standard Enthalpy of Formation of Gaseous Water Is -241