Standard enthalpy of vaporization Δ H ° for water at 100° C is 40.66 kJ / mol -1 . The internal energy change of vaporization of water at 100° C (in kJ / mol -1 ) is:

Question

Standard enthalpy of vaporization Δ H ° for water at 100° C is 40.66 kJ / mol -1 . The internal energy change of vaporization of water at 100° C (in kJ / mol -1 ) is: (A) 37.56 (B) -43.16 (C) +43.76 (D) +40.66

Tardigrade Admin · Accepted Answer

H 2 O (ℓ) arrow H 2 O ( g ) Δ H =Δ E +Δ n g RT 40.66=Δ E +1 × (8.314/1000) × 373 40.66=Δ E +3.101 Δ E =37.56

Q. Standard enthalpy of vaporization $Δ H^{\circ}$ for water at $10 0^{\circ} C$ is $40.66 k J / m o l^{- 1} .$ The internal energy change of vaporization of water at $10 0^{\circ} C$ (in $k J / m o l^{- 1}$ ) is:

37.56

-43.16

+43.76

+40.66

$H_{2} O (ℓ) \to H_{2} O (g)$

$Δ H = Δ E + Δ n_{g} RT$

$40.66 = Δ E + 1 \times \frac{8.314}{1000} \times 373$

$40.66 = Δ E + 3.101$

$Δ E = 37.56$

Q. Standard enthalpy of vaporization ΔH∘ for water at 100∘C is 40.66kJ/mol−1. The internal energy change of vaporization of water at 100∘C (in kJ/mol−1 ) is: