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, OH_fus= 6.01 kJ/mol, and OH_vap= 40.67 kJ/mol A) 12.28 B) 13.16 C) 6.41 D) 9.40 E) 7154

Question

Quizplus · Accepted Answer

The enthalpy change can be calculated by breaking down the process into several steps: 
1. Heating the ice from -50.0°C to its melting point at 0°C: q = nCΔT = (1.00 mol)(2.09 J/g-K)(50.0°C) = 522.5 J = 0.5225 kJ
2. Melting the ice at 0°C: ΔHfus = 6.01 kJ/mol
3. Heating the water from 0°C to 70.0°C: q = nCΔT = (1.00 mol)(4.18 J/g-K)(70.0°C) = 293.2 J = 0.2932 kJ
4. Heating the water vapor from 100°C to 70.0°C: q = nCΔT = (1.00 mol)(1.84 J/g-K)(70.0°C - 100°C) = -55.2 J = -0.0552 kJ (negative sign indicates that the water vapor is losing heat)
5. Condensing the water vapor at 100°C: ΔHvap = -40.67 kJ/mol (negative sign indicates that energy is being released)

The overall enthalpy change is the sum of these individual changes: 
ΔH = q1 + ΔHfus + q2 + q3 + ΔHvap = 0.5225 kJ + 6.01 kJ/mol + 0.2932 kJ - 0.0552 kJ - 40.67 kJ/mol = -28.93 kJ/mol
The answer is rounded to two decimal places, giving choice B.

The Enthalpy Change for Converting 1