**Changes of State and ΔH°**

Calculate the energy transferred to the surroundings when water vapor in the air condenses at 25 °C to give rain in a thunderstorm. Suppose that one inch of rain falls over one square mile of ground, so that 6.6 × 10^{10} mL has fallen. (Assume d_{H_2O(\ell)} = 1.0 g/mL.)

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1.6 × 10^{11} kJ

**Strategy and Explanation** The thermochemical expression for condensation of 1 mol water at 25 °C is

H_2O(g) → H_2O(\ell) ΔH° = – 44.0 kJ

The standard enthalpy change tells how much heat transfer is required when 1 mol water condenses at constant pressure, so we first calculate how many moles of water condensed.

Amount of water condensed = 6.6 × 10^{10} g water × \frac{1 mol}{18.0 g} = 3.66 × 10^9 mol water

Next, calculate the quantity of energy transferred from the fact that 44.0 kJ is transferred per mole of water.

Quantity of energy transferred = 3.66 × 10^9 mol water × \frac{44.0 kJ}{1 mol} = 1.6 × 10^{11} kJ

The negative sign of ΔH° in the thermochemical expression indicates transfer of the 1.6 × 10^{11} kJ from the water (system) to the surroundings.

**Reasonable Answer Check ** The quantity of water is about 10^{11} g. The energy transfer is 44 kJ for 1 mol (18 g) water. Since 44 is about twice 18, this is about 2 kJ/g. Therefore, the number of kJ transferred in kJ should be about twice the number of grams, or about 2 × 10^{11} kJ, and it is.

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