Calculate the enthalpy change for converting 2.00 mol of ice at -50.0°C to water at 70.0°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 2.00 mol of ice at -50.0°C to water at 70.0°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

-50.0°C → 0°C: ice

0°C → 70.0°C: water

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

ΔH_fus (H₂O) = 6.01 kJ.mol^-1

ΔH_fus (H₂O) (in J) = ΔH_fus (H₂O) (in J.mol^-1) x n_H2O

ΔH_fus (H₂O) (in J) = 6.01 x 10³ x 2 = 1.20 x 10⁴ J

n =

\frac{m}{M}

⇒ m_H2O= n_H2O x M_H2O= 2.00 x 18.0 = 36.0 g

ΔH = ΔH_fus (H₂O) (in J) + q_P (ice) (in J) + q_P (water) (in J)

ΔH = ΔH_fus (H₂O) + m ΔT c_S (ice) + m ΔT c_S (water)

ΔH = 1.20 x 10⁴ + 36.0 x (273-223) x 2.09 + 36.0 x (343-273) x 4.18

ΔH = 26.3 x 10⁴ J = 26.3 kJ

Exercise 4 | Liquids and Solids