Calculate the enthalpy change (in kJ)associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 109.0 °C.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,Δ = 6.01 kJ/mol and Δ = 40.67 kJ/mol. A)64.8 B)75.9 C)11100 D)12000 E)112

Question

Quizplus · Accepted Answer

The conversion of ice at -4 °C to water at 0 °C requires the following heat:
Q1 = m * Cice * ΔT = 25.0 g * 2.09 J/g-K * 4.00 K = 209 J
The conversion of water at 0 °C to water at 100 °C requires the following heat:
Q2 = m * Cwater * ΔT = 25.0 g * 4.18 J/g-K * 100.00 K = 10450 J
The conversion of water at 100 °C to water vapor at 109 °C requires the following heat:
Q3 = m * ΔHvap + m * Csteam * ΔT = 25.0 g * (40.67 kJ/mol / 18.02 g/mol) + 25.0 g * 1.84 J/g-K * 9.00 K = 29623 J
The total heat required for the conversion is:
ΔH = Q1 + Q2 + Q3 = 209 J + 10450 J + 29623 J = 40282 J = 40.28 kJ
Therefore, the enthalpy change is 40.28 kJ, which is closest to 75.9 kJ (answer B).

Calculate the Enthalpy Change (In KJ)associated with the Conversion of 25.0