General Chemistry I

Question

Calculate the enthalpy change converting 2.5g of water at -35.0 C to steam at 140.0 C under constant pressure of 1 atm. The specific heats of ice, liquid water, and steam are 2.03J/g-K, 4.18 J/g-K, and 1.84 J/g-k respectively. For H2O, triangle h fus = 6.01 kJ/mol and triangle h vap = 40.67 kJ/mol.

Answer 1

For these types of problems, it is best to do them in steps.

We will use values of heat (q), mass (m), change in temperature (∆T), specific heat (C), heat of fusion (∆Hf) and heat of vaporization (∆Hvap)

step 1: convert 2.5 g ice @ -35.0º to ice at 0º

q = mC∆T = (2.5 g)(2.03 J/gº)(35º) = 178 J

step 2: convert 2.5 g of ice at 0º to liquid water at 0º (a phase change, no change in temperature)

q = m∆Hf

converting 6.01 kJ/mol to J/g we have 6.01 kJ/mol x 1000 J/kJ x 1 mol / 18 g = 334 J/g

q = (2.5 g)(334 J/g) = 835 J

step 3: raise temp of 2.5 g water from 0º to 100º

q = mC∆T = (2.5 g)(4.18 J/gº)(100º) = 1045 J

step 4: convert 2.5 g water @100º to steam @100º (phase change, no change in temperature)

converting 40.67 kJ/mol to J/g we have 40.67 kJ/mol x 1000 J/kJ x 1 mol / 18 g = 2259 J/g

q = m∆Hvap = (2.5 g)(2259 J/g) = 5648 J

step 5: raise the temp of 2.5 g steam from 100º to 140º

q = mC∆T = (2.5 g)(1.84 J/gº)(40º) = 184 J

step 6: add up all the heat values

178 J + 835 J + 1045 J + 5648 J + 184 J = 7890 J

So the enthalpy change is 7890 J or 7.89 kJ