Please help, I am confused about this problem?

Question

Okay here is the problem

When it rains gas phase water in the atmosphere must be converted to liquid phase water (i.e. condensation.: H2O(g) -> H2O (l) ). Calculate the standard change in enthalpy associated with a thunderstorm in which 1.0 inch of rain falls over an area of 1.0 mile^2. List any human related events which have taken place on earth which are of comparable energy.

Please help I am really confused about this problem and would appreciate any help.

Answer 1

First convert your Imperial measurements to metric, using 1 inch = 2.54 cm.

In 1 mile there are 63360 inches.
63360 inches x 2.54 = 160934.4 cm
In 1 sq mile there are 160934.4^2 cm^2
In 1 sq mile by 1 inch deep there are 160934.4^2 x 2.54 cm^3
160934.4^2 x 2.54 = 6.58 x 10^10 cm^3

NB Because numbers are becoming so large we now resort to scientific notation.

6.58 x 10^10 cm^3 = 6.58 x 10^10 g (grams).

NB It is a standard that 1 cm^3 of water = 1 gram of water.

Calculate moles of water
moles(H2O) = 6.58 x 10^10 / 18 (Mr of water)
moles(H20) = 3.655 x 10^9 moles(water)

The enthalpy change when water changes from gas phase to liquid phase it releases -44kJmol^-1 of energy.
That is -44 kJ for 1 mole.
but we have 3.655 x 10^9 moles
So total energy release is -44 x 3.655 x 10^9 = = -1.608 x 10^11 kJ is released.
So the standard enthalpy change is -1.608 x 10^11 kJ.

A human related event that may release such an amount of energy is launching a rocket into space.

Answer 2

Enthalpy of change for water is 40.65 kJ/mol for vaporization

You know how much water as a volume fell (1in * 1 mi^2)

You need to know the density of water... 0.998 g/cm^3 (@20C)

Then you know the molar mass of water is 18.0153 g/mol

Change from volume to weight to molar mass. Then multiply by the enthalpy of change.

(There is also a way to do it with the molarity of water being ~55 mol/L but either way you get to the answer)