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, ΔH_fus = 6.01 kJ/mol and ΔH_vap = 40.67 kJ/mol. A)64.8 B)75.9 C)11100 D)12000 E)112

Question

Quizplus · Accepted Answer

The process of converting ice to water vapor involves several steps and each step contributes to the total enthalpy change. We can break down the process into the following steps:

1. Heating the ice from -4.00 °C to 0.00 °C and melting it to water at 0.00 °C.
2. Heating the water from 0.00 °C to 100.00 °C and vaporizing it to steam at 100.00 °C.
3. Heating the steam from 100.00 °C to 109.00 °C.

To calculate the enthalpy change for each step, we need to use the following formulas:

q = mCΔT (for heating or cooling)
q = nΔH (for melting or vaporization)

where q is the heat absorbed or released, m is the mass of the substance, C is the specific heat of the substance, ΔT is the change in temperature, n is the moles of the substance, and ΔH is the enthalpy of melting or vaporization.

Step 1: Heating and melting the ice

q = (25.0 g)(2.09 J/g-K)(0.00 °C - (-4.00 °C)) = 209 J (heating)
q = (25.0 g)/(18.015 g/mol)(6.01 kJ/mol) = 20.8 kJ (melting)

Step 2: Heating and vaporizing the water

q = (25.0 g)(4.18 J/g-K)(100.00 °C - 0.00 °C) = 10,450 J (heating)
q = (25.0 g)/(18.015 g/mol)(40.67 kJ/mol) = 58.1 kJ (vaporization)

Step 3: Heating the steam

q = (25.0 g)(1.84 J/g-K)(109.00 °C - 100.00 °C) = 407 J (heating)

The total enthalpy change is the sum of the enthalpy changes for each step:

ΔH = 20.8 kJ + 58.1 kJ + 407 J = 79.3 kJ

The answer is closest to choice B: 75.9 kJ. However, this is not an option, so the best choice is B. The answer is positive because energy is absorbed by the system (ice) to overcome the intermolecular forces and change its state to water vapor.

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