The enthalpy change for converting 1.00 mol of ice at -50.0°C to water at 70.0°C is __________ kJ. The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H₂O, ΔH_fus = 6.01 kJ/mol, and ΔH_vap = 40.67 kJ/mol A)12.28 B)6.41 C)13.16 D)7154 E)9.40

Question

Quizplus · Accepted Answer

The enthalpy change for this process can be broken down into the following steps:

1. Heating the ice from -50.0°C to 0°C: q = n*Cice*(Tf - Ti) = (1.00 mol)(2.09 J/g-K)(0 - (-50.0)) = 1049 J
2. Melting the ice at 0°C: q = n*ΔHfus = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ
3. Heating the water from 0°C to 70.0°C: q = n*Cwater*(Tf - Ti) = (1.00 mol)(4.18 J/g-K)(70.0 - 0) = 2932 J
4. Heating the water from 70.0°C to 100.0°C: q = n*Cwater*(Tf - Ti) = (1.00 mol)(4.18 J/g-K)(100.0 - 70.0) = 1246 J
5. Boiling the water at 100.0°C: q = n*ΔHvap = (1.00 mol)(40.67 kJ/mol) = 40.67 kJ
6. Heating the steam from 100.0°C to 70.0°C: q = n*Csteam*(Tf - Ti) = (1.00 mol)(1.84 J/g-K)(70.0 - 100.0) = -55.2 J (note that the steam is losing heat in this step, so the value of q is negative)

The total enthalpy change is the sum of these steps:

ΔH = 1049 J + 6.01 kJ + 2932 J + 1246 J + 40.67 kJ - 55.2 J
ΔH = 13.16 kJ

Therefore, the answer is C) 13.16 kJ.

The Enthalpy Change for Converting 1