Calculate the enthalpy change for converting 10.0 g of water at 25.0°C to steam at 135.0°C.

Data: c_S (ice) = 2.09 J.g^-1.K^-1; c_S (water) = 4.18 J.g^-1.K^-1; c_S (steam) = 1.84 J.g^-1.K^-1

ΔH_fus (H₂O) = 6.01 kJ.mol^-1; ΔH_vap (H₂O) = 40.67 kJ.mol^-1

Question

Calculate the enthalpy change for converting 10.0 g of water at 25.0°C to steam at 135.0°C.

Data: c_S (ice) = 2.09 J.g^-1.K^-1; c_S (water) = 4.18 J.g^-1.K^-1; c_S (steam) = 1.84 J.g^-1.K^-1

ΔH_fus (H₂O) = 6.01 kJ.mol^-1; ΔH_vap (H₂O) = 40.67 kJ.mol^-1

| General Chemistry 2

Accepted Answer

25.0°C → 100.0°C: water

100.0°C → 135.0°C: steam

c_S = $\frac{q_{P}}{m ∆ T}$ with q_P = energy added as heat (in J)

ΔH_vap (H₂O) = 40.67 kJ.mol^-1

10.0 g of water ⇒ 5.56 x 10^-1 mol of water

So, for this reaction ΔH_vap (H₂O) (in J) = 40.67 x 0.556 = 22.6 kJ

ΔH = ΔH_vap (H₂O) (in J) + q_P (water) (in J) + q_P (steam) (in J)

ΔH = ΔH_vap (H₂O) + mΔT c_S (water) + mΔT c_S (steam)

ΔH = 22.6 x 10³ + 10.0 x (373-298) x 4.18 + 10.0 x (408-373) x 1.84

ΔH = 26.4 x 10³ J = 26.4 kJ

Exercise 1 | Liquids and Solids