One mole of H2O(g) at 1.00 atm and 100 degrees C occupies a volume of 30.6 L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00 atm and 100 degrees C, 40.66 kJ of heat is released.
If the density of H2O(l) at this temperature and pressure is 0.996 g/cm(cubed), calculate that change of enthalpy for the condensation of one mole of water at 1.00 atm and 100 degrees C
To calculate the change of enthalpy for the condensation of one mole of water at 1.00 atm and 100 degrees Celsius, we can use the equation:
ΔH = q / n
Where:
ΔH is the change of enthalpy,
q is the heat released or absorbed,
n is the number of moles.
Given that 40.66 kJ of heat is released and we are condensing one mole of water, we can write:
q = -40.66 kJ
n = 1 mol
Substituting these values into the formula, we get:
ΔH = -40.66 kJ / 1 mol
Now, we need to convert the units of kJ to J since the SI unit for enthalpy is joules (J).
1 kJ = 1000 J, so:
ΔH = -40.66 kJ * 1000 J/kJ / 1 mol
ΔH = -40,660 J / 1 mol
Finally, to convert the change in enthalpy to grams, we need to use the density of water. The density of water is given as 0.996 g/cm^3.
Since the volume of one mole of water is 30.6 L, we need to convert the volume to cm^3:
1 L = 1000 cm^3
30.6 L * 1000 cm^3/L = 30,600 cm^3
Since the density is given in g/cm^3, we can calculate the mass of one mole of water:
mass = density * volume
mass = 0.996 g/cm^3 * 30,600 cm^3 = 30,417.6 g
Now we can determine the change of enthalpy per gram of water:
ΔH/g = ΔH / mass
ΔH/g = -40,660 J / 30,417.6 g
Calculating this, we find:
ΔH/g ≈ -1.337 J/g