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Constants for Water:

[tex]\[
\begin{array}{|l|l|}
\hline
\Delta H_{\text{vap}} & 40.65 \, \text{kJ/mol} \\
\hline
\Delta H_f & -285.83 \, \text{kJ/mol} \\
\hline
\Delta H_{\text{fusion}} & 6.03 \, \text{kJ/mol} \\
\hline
\text{Specific Heat} & 4.186 \, \text{J/g}^\circ\text{C} \\
\hline
\text{Molar Mass} & 18.02 \, \text{g/mol} \\
\hline
\end{array}
\][/tex]

How much energy is generated from freezing 2.5 g of water?

A. [tex]\(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 4.186 \, \text{kJ/mol}\)[/tex]

B. [tex]\(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 6.03 \, \text{kJ/mol}\)[/tex]

C. [tex]\(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 40.65 \, \text{kJ/mol}\)[/tex]

D. [tex]\(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 285.83 \, \text{kJ/mol}\)[/tex]

---

(Note: The correct answer should be option B based on the enthalpy of fusion for freezing water.)



Answer :

GinnyAnswer
To determine the energy generated from freezing [tex]\(2.5 \text{ g}\)[/tex] of water, you need to follow these steps:

1. Determine the number of moles of water:
- Given the mass of water [tex]\(2.5 \text{ g}\)[/tex] and its molar mass [tex]\(18.02 \text{ g/mol}\)[/tex], calculate the number of moles using the formula:
[tex]\[ \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of water} = \frac{2.5 \text{ g}}{18.02 \text{ g/mol}} \approx 0.1387 \text{ mol} \][/tex]

2. Calculate the energy generated from freezing:
- The enthalpy change for the phase transition from liquid to solid (freezing point) is known as the enthalpy of fusion ([tex]\(\Delta H_{\text{fusion}}\)[/tex]), which is given as [tex]\(6.03 \text{ kJ/mol}\)[/tex].
- The energy generated can be calculated by multiplying the number of moles of water by the enthalpy of fusion:
[tex]\[ \text{energy generated} = \text{moles of water} \times \Delta H_{\text{fusion}} \][/tex]
Plugging in the values:
[tex]\[ \text{energy generated} = 0.1387 \text{ mol} \times 6.03 \text{ kJ/mol} \approx 0.8366 \text{ kJ} \][/tex]

Therefore, the correct option is:

B. [tex]\(2.5 \text{ g} \times \frac{1 \text{ mol}}{18.02 \text{ g}} \times 6.03 \text{ kJ/mol}\)[/tex]

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