Question

The enthalpy change when 1 g of water is frozen at 0°C ($\Delta_{fus}H^0$ = 1.435 kcl/mol) is

Answer 1

-0.08 kcal

Answer 2

1. Convert grams to moles:
   Moles of water = $\frac{1\,\text{g}}{18\,\text{g/mol}} \approx 0.0556\,\text{mol}$.

2. Calculate enthalpy change (freezing):
   For freezing, the enthalpy change is the negative of the fusion value.

   $$\Delta H = -0.0556\,\text{mol} \times 1.435\,\text{kcal/mol} \approx -0.0797\,\text{kcal}$$

   This approximately equals $-0.08\,\text{kcal}$.

Core Explanation:
Convert 1 g water to moles (1/18 mol) and multiply by $\Delta_{fus}H^0$, then take the negative sign because freezing is exothermic.