The enthalpy change for converting 1.00 mol of ice at -60.0°C to water at 80.0°C is The specific heats of ice, water, and steam are and respectively. For O, = 6.01 kJ mol^-1, and . A) 9.77 B) 14.29 C) 8282 D) 13.17 E) 48.95

Question

Quizplus · Accepted Answer

The enthalpy change for the conversion can be broken down into three steps.

Step 1: Convert ice at -60.0°C to ice at 0°C.

q1 = m × ΔHfus(ice) = (1.00 mol) × (6.01 kJ mol^-1) = 6.01 kJ

Step 2: Convert ice at 0°C to water at 0°C.

q2 = m × Cp(ice) × ΔT + m × ΔHfus(water) = (1.00 mol) × (37.1 J K^-1 mol^-1) × (0°C - (-60.0°C)) + (1.00 mol) × (6.01 kJ mol^-1) = 9.77 kJ

Step 3: Convert water at 0°C to water at 80.0°C.

q3 = m × Cp(water) × ΔT = (1.00 mol) × (75.3 J K^-1 mol^-1) × (80.0°C - 0°C) = 2.40 kJ

Therefore, the total enthalpy change is:

ΔH = q1 + q2 + q3 = 6.01 kJ + 9.77 kJ + 2.40 kJ = 18.18 kJ

The closest answer choice to 18.18 kJ is 14.29 kJ (B).

The Enthalpy Change for Converting 1