Calculate the enthalpy change, ÄH DeltaH, for the process in which 12.7g g of water is converted from liquid at 7.3 ∘ C \, ^\circ C to vapor at 25.0 ∘ C \, ^\circ C .
For water, ÄH vap \Delta H_{\rm vap} = 44.0kJ/mol kJ/mol at 25.0 ∘ C \, ^\circ C and s s = 4.18J/(g⋅ ∘ C) J/(g\cdot{ ^\circ C}) for H 2 O(l)
dH for H2O moving from 7.3C to 25C is
dH = mass H2O x specific heat H2O x (Tfinal-Tinitial)
dH for vaporization at 25C is
dH = mass H2O x Hvap.
Add the two together for the total.
To calculate the enthalpy change (ΔH) for the process, you need to consider two steps: heating the water from 7.3°C to its boiling point and then vaporizing it at its boiling point to reach 25.0°C.
Step 1: Heating the water:
First, calculate the heat energy required to raise the temperature of 12.7g of water from 7.3°C to its boiling point. The specific heat capacity (s) of water is given as 4.18J/(g⋅°C).
Q1 = m * s * ΔT1
where:
Q1 = heat energy
m = mass of water = 12.7g
s = specific heat capacity of water = 4.18J/(g⋅°C)
ΔT1 = change in temperature = boiling point - initial temperature = 100°C - 7.3°C
Step 2: Vaporizing the water:
Secondly, calculate the heat energy required to vaporize the water. The enthalpy of vaporization (ΔHvap) for water is given as 44.0 kJ/mol at 25.0°C. We need to convert the mass of water to molar mass to use this value.
First, calculate the number of moles of water:
moles = mass / molar mass
The molar mass (Mw) of water is approximately 18.015 g/mol.
Now, calculate the heat energy to vaporize the water using the enthalpy of vaporization:
Q2 = moles * ΔHvap
Finally, calculate the total enthalpy change (ΔH) by summing up Q1 and Q2:
ΔH = Q1 + Q2
Plug in all the given values and calculate to find the enthalpy change.