Question

Calculate the enthalpy change upon converting 1.00 mol of ice at -25 °C to steam at 125 °C under a constant pressure of 1 atm. The specific heats of ice, liquid water, and steam are 2.03, 4.18, and 1.84 J/g-K, respectively. For H₂O, ΔH_fus = 6.01 kJ/mol and ΔH_vap=40.67 kJ/mol. 

Answer 1

The process of changing ice to steam involves changes in phases. The process involves following steps:

1. Heating the ice to 0⁰C.

2. Melting of ice to water at 0⁰C.

3. Heating the water from 0⁰C to 100⁰C.

4. Vaporization of water at 100⁰C.

5. Heating the vapor to 125⁰C.

Calculate the heat required for each of the steps and then sum them up to get the total enthalpy change for the process.

Use the following formula to calculate the heat required:

where, q is heat energy, c is specific heat and is change in temperature.

1. Heating the ice from -25⁰C to 0⁰C .

q₁ = (1.00 mol*18.0 g/mol) * 2.03 J/g.K * [0⁰C-(-25⁰C)]

= 913.5 J = 0.9135 KJ

2. Conversion of ice to water at 0⁰C.

Here, we can use heat of fusion as the phase changes from solid to liquid.

q₂ = 1.00 mol * 6.01 KJ/mol = 6.01 KJ

3. Heating the water from 0⁰C to 100⁰C.

q₃ = (1.00 mol*18.0 g/mol) * 4.18 J/g.K * 100 K

= 7524 J = 7.524 KJ

4. conversion of water to steam at 100⁰C.

Here, use the enthalpy of vaporization as the there is a phase change from liquid to vapor.

q₄ = 1.00 mol * 40.67 KJ/mol = 40.67 KJ

5. Heating of vapor to 125⁰C.

q₅ = (1.00 mol*18.0 g/mol) * 1.84 J/g.K * 25 K

= 828 J = 0.828 KJ

Now, calculate the total enthalpy change.

= 0.9135 KJ + 6.01 KJ + 7.524 KJ + 40.67 KJ + 0.828 KJ

= 55.9 KJ