Answer :
To determine how much energy is generated from freezing \(2.5 \, \text{g}\) of water, we will proceed step-by-step.
1. Identify the relevant constants:
For freezing water, the relevant enthalpy change is the heat of fusion, \(\Delta H_{\text{fusion}} = 6.03 \, \text{kJ/mol}\).
2. Determine the mass of water:
Given mass of water \( m = 2.5 \, \text{g} \).
3. Calculate the moles of water:
We need to convert the mass of water to moles. The molar mass of water, \( M_{\text{H}_2\text{O}} = 18.02 \, \text{g/mol} \).
[tex]\[ \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{2.5 \, \text{g}}{18.02 \, \text{g/mol}} = 0.1387 \, \text{mol} \][/tex]
4. Calculate the energy using \(\Delta H_{\text{fusion}}\):
[tex]\[ \text{energy generated} = (\text{moles of water}) \times \Delta H_{\text{fusion}} \][/tex]
[tex]\[ \text{energy generated} = 0.1387 \, \text{mol} \times 6.03 \, \text{kJ/mol} = 0.8366 \, \text{kJ} \][/tex]
Therefore, the amount of energy generated from freezing \(2.5 \, \text{g}\) of water is \(0.8366 \, \text{kJ}\).
The closest matching option in your list is:
D. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 6.03 \, \text{kJ/mol} \)
So, the correct choice is:
D. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times (6.03) \, \text{kJ/mol}
1. Identify the relevant constants:
For freezing water, the relevant enthalpy change is the heat of fusion, \(\Delta H_{\text{fusion}} = 6.03 \, \text{kJ/mol}\).
2. Determine the mass of water:
Given mass of water \( m = 2.5 \, \text{g} \).
3. Calculate the moles of water:
We need to convert the mass of water to moles. The molar mass of water, \( M_{\text{H}_2\text{O}} = 18.02 \, \text{g/mol} \).
[tex]\[ \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{2.5 \, \text{g}}{18.02 \, \text{g/mol}} = 0.1387 \, \text{mol} \][/tex]
4. Calculate the energy using \(\Delta H_{\text{fusion}}\):
[tex]\[ \text{energy generated} = (\text{moles of water}) \times \Delta H_{\text{fusion}} \][/tex]
[tex]\[ \text{energy generated} = 0.1387 \, \text{mol} \times 6.03 \, \text{kJ/mol} = 0.8366 \, \text{kJ} \][/tex]
Therefore, the amount of energy generated from freezing \(2.5 \, \text{g}\) of water is \(0.8366 \, \text{kJ}\).
The closest matching option in your list is:
D. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 6.03 \, \text{kJ/mol} \)
So, the correct choice is:
D. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times (6.03) \, \text{kJ/mol}
