Question

Given the thermochemical expression \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\) $$ \Delta_{\mathrm{r}} H^{\circ}=6.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate what quantity of energy is transferred to the surroundings when (a) 34.2 mol liquid water freezes. (b) \(100.0 \mathrm{~g}\) liquid water freezes.

Short answer

(a) 205.2 kJ; (b) 33.3 kJ.

Step-by-step solution

Understand the Given Reaction

The given reaction, \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\), represents the melting of ice to form liquid water, with \(\Delta_{\mathrm{r}} H^{\circ} = 6.0 \, \mathrm{kJ/mol}\). When the reverse occurs, freezing (from liquid to solid), \(\Delta_{\mathrm{r}} H^{\circ}\) will be \(-6.0 \, \mathrm{kJ/mol}\), indicating energy is released.

Calculate Energy for 34.2 mol of Liquid Water Freezing

To find the energy transferred to the surroundings when 34.2 mol of liquid water freezes, use the enthalpy change for freezing, which is \(-6.0 \, \mathrm{kJ/mol}\). This is calculated by multiplying the number of moles by the enthalpy change: \[\text{Energy} = 34.2 \, \text{mol} \times (-6.0 \, \mathrm{kJ/mol}) = -205.2 \, \mathrm{kJ}.\] Thus, 205.2 kJ of energy is released to the surroundings.

Convert Mass to Moles for 100.0 g of Water

First, convert 100.0 grams of liquid water to moles. Using the molar mass of water \(\mathrm{(18.02 \, g/mol)}\), calculate the moles as follows: \[\text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{100.0 \, \mathrm{g}}{18.02 \, \mathrm{g/mol}} = 5.55 \, \text{mol}.\]

Calculate Energy for 5.55 mol of Liquid Water Freezing

Now that we have the moles, calculate the energy released when 5.55 mol of liquid water freezes. Multiply the number of moles by the enthalpy change: \[\text{Energy} = 5.55 \, \text{mol} \times (-6.0 \, \mathrm{kJ/mol}) = -33.3 \, \mathrm{kJ}.\] Thus, 33.3 kJ of energy is released to the surroundings.

Key concepts

Enthalpy Change

Enthalpy change is a key concept in thermochemistry, representing the heat absorbed or released during a chemical reaction at constant pressure.
In the given exercise, we deal with changes in the physical state of water, specifically melting and freezing. For these phase changes, enthalpy changes tell us about energy transfer.
- **Positive Enthalpy Change**: This happens for endothermic processes, where energy is absorbed. For melting ice, the enthalpy change, \( \Delta_{\text{r}} H^{\circ} = 6.0 \, \text{kJ/mol} \), indicates energy is absorbed from the environment to turn ice into liquid water. - **Negative Enthalpy Change**: This occurs in exothermic processes, where energy is released. When water freezes, the reverse of the melting process, \( \Delta_{\text{r}} H^{\circ} = -6.0 \, \text{kJ/mol} \), showing that energy is released to the surroundings. Enthalpy changes not only guide us in calculating energy transfers but also indicate whether processes are spontaneous or need energy input.

Phase Transitions

Phase transitions refer to changes between different states of matter: solid, liquid, and gas. For water, these transitions are familiar yet fascinating physical processes. - **Melting (Solid to Liquid)**: This is when ice melts to form liquid water. Energy must be absorbed from the surroundings to overcome the molecular forces holding water molecules in a solid structure. - **Freezing (Liquid to Solid)**: This is the reverse of melting. When water freezes, energy is released back to the environment, because the molecules form a more ordered solid structure, releasing energy as they're bound more tightly. Understanding these transitions involves recognizing the enthalpy changes. During both melting and freezing, the temperature doesn't change even though energy is absorbed or released. This is because the energy change is used in breaking or forming the hydrogen bonds, not altering temperature.

Energy Calculations

Energy calculations in thermochemistry often involve determining how much energy is transferred during reactions or phase changes. The steps often include understanding the relevant enthalpy change and applying it to the quantities given. In our example:

• To calculate the energy for 34.2 mol of water freezing, we took the moles and multiplied by the enthalpy change for freezing (which is negative due to energy release) to find the total energy released, \(34.2 \, \text{mol} \times (-6.0 \, \text{kJ/mol}) = -205.2 \, \text{kJ}\).
• When dealing with 100.0 grams of water, the mass was first converted to moles. Using the molar mass of water, 18.02 g/mol, we calculated the moles and then used the enthalpy change to find the energy released, \(5.55 \, \text{mol} \times (-6.0 \, \text{kJ/mol}) = -33.3 \, \text{kJ}\).

Energy calculations are crucial because they allow us to predict and measure the heat exchange involved in physical and chemical processes, thus understanding the dynamics of reactions and transitions in detail.