# Question 6.PS.6: Changes of State and ΔH° Calculate the energy transferred to......

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 $d_{H_2O(\ell)} = 1.0 g/mL.)$

Step-by-Step
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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.

Question: 6.PS.14

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Question: 6.PS.13

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Question: 6.PS.12

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Question: 6.PS.11

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Question: 6.PS.10

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Question: 6.PS.9

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Question: 6.PS.8

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Question: 6.PS.7

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Question: 6.PS.5

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