# Question #557f1

Apr 1, 2015

Hess's Law states that the enthalpy change for a reaction can be expressed as the sum of enthalpy changes for several individual steps. Therefore, if you're asked to determine the enthalpy change for a reaction under non-standard conditions, and you only have a table for standard conditions, you can construct a more complicated pathway and simply sum the enthalpy changes for the individual steps.

Example:

Determine the enthalpy change for cooling 125 g of water from 25 C to -15 C.

This process can be broken down into 3 steps:
1) Cooling liquid water from 25 to 0 C
2) Freezing the liquid to solid ice
3) Cooling the ice from 0 to -15 C

1: The specific heat capacity of liquid water is about 4.18 J/(g-K), so the enthalpy change of the first step is
$\Delta {H}_{1} = \left(125 g\right) \left(4.18 \frac{J}{g - K}\right) \left(0 - 25\right) = - 13062 J$

2: The enthalpy of fusion of water is 333.5 J/g, so the enthalpy change of the second step is
$\Delta {H}_{2} = \left(125 g\right) \left(- 333.5\right) = - 41687 J$

3: The specific heat capacity of ice is about 2.03 J/(g-K), so the enthalpy change of the third step is
$\Delta {H}_{3} = \left(125 g\right) \left(2.03 \frac{J}{g - K}\right) \left(- 15 - 0\right) = - 3806 J$

The total enthalpy change of the process is the sum of the individual steps:
$\Delta {H}_{T o t a l} = - 13 , 062 J - 41 , 687 J - 3 , 806 J = - 58 , 555 J$