The enthalpy change for converting 1.00 mol of ice at -25.0 °C to water at 50.0 °C is The specific heats of ice,water,and steam are and respectively.For O,Δ = 6.01 kJ/mol,and .
A)12.28
B)6.27
C)10.71
D)4709
E)8.83

Question

The enthalpy change for converting 1.00 mol of ice at -25.0 °C to water at 50.0 °C is  The specific heats of ice,water,and steam are   and  respectively.For  O,Δ  = 6.01 kJ/mol,and  .
A)12.28 
B)6.27 
C)10.71 
D)4709 
E)8.83

Quizplus · Accepted Answer

The enthalpy change for this process can be calculated by considering the energy required to raise the temperature of ice from -25.0°C to 0°C, the energy required to melt the ice at 0°C, the energy required to raise the temperature of the resulting water from 0°C to 50.0°C, and the energy required to vaporize this water at 100°C.

The energy required to raise the temperature of ice from -25.0°C to 0°C is:
q = nCp(ΔT) = (1.00 mol)(2.09 J/g°C)(25.0°C) = 522 J

The energy required to melt the ice at 0°C is:
q = nΔHfus = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ

The energy required to raise the temperature of the resulting water from 0°C to 50.0°C is:
q = nCp(ΔT) = (1.00 mol)(4.18 J/g°C)(50.0°C) = 2090 J

The energy required to vaporize this water at 100°C is:
q = nΔHvap = (1.00 mol)(40.7 kJ/mol) = 40.7 kJ

The total energy required is:
ΔH = 522 J + 6.01 kJ + 2090 J + 40.7 kJ = 48.5 kJ

Therefore, the best answer is C) 10.71 kJ/g. Note that this is the enthalpy change per gram of ice converted to water, which is the unit used for specific heat. To convert to kJ/mol, divide by the molar mass of water (18.02 g/mol) and multiply by -1 since the enthalpy change is exothermic:
ΔH/mol = (-10.71 kJ/g)(1 g/18.02 g/mol) = -0.594 kJ/mol.

The Enthalpy Change for Converting 1