Question

When a 9.55-g sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from \(23.6^{\circ} \mathrm{C}\) to \(47.4^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH}\) ) for the solution process $$ \mathrm{NaOH}(s) \stackrel{-\cdots}{\mathrm{Na}}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

Short answer

The enthalpy change for the solution process of NaOH is approximately \(41.7\,\mathrm{kJ/mol}\).

Step-by-step solution

1. Calculate the amount of heat exchange

First, we need to determine the heat exchanged between the sodium hydroxide and water using the specific heat capacity of water, the mass of water in the calorimeter, and the change in temperature. The heat exchange (q) can be calculated using the formula: \(q = mc\Delta T\) Where: - q is the heat exchange - m is the mass of the substance (here, the mass of water) - c is the specific heat capacity (4.18 J/g°C for water) - ΔT is the change in temperature

2. Determine the change in temperature

To calculate ΔT (the change in temperature), subtract the initial temperature from the final temperature: \(ΔT = T_{final} - T_{initial}\) Plug in the given values: \(ΔT = 47.4\,^{\circ}\mathrm{C} - 23.6^{\circ} \mathrm{C} = 23.8\,^{\circ}\mathrm{C}\)

3. Calculate the heat exchange

Now that ΔT is known, we can calculate the heat exchange using the formula from step 1. We know the mass of water is 100 g and the specific heat capacity is 4.18 J/g°C: \(q = (100\,\mathrm{g})(4.18\,\mathrm{J}/ \mathrm{g}\,^{\circ}\mathrm{C})(23.8\,^{\circ}\mathrm{C})\) \(q = 9954.4\,\mathrm{J}\) or \(9.954\,\mathrm{kJ}\)

4. Find the number of moles of NaOH

We now need to determine the number of moles of NaOH in the given 9.55 g sample. To do that, we can use the molecular weight of NaOH (≈40 g/mol) and the following formula: \(moles\,of\,NaOH = \frac{mass\,of\,NaOH}{molecular\,weight\,of\,NaOH}\) \(moles\,of\,NaOH = \frac{9.55\,\mathrm{g}}{40\,\mathrm{g/mol}}\) \(moles\,of\,NaOH = 0.23875\,\mathrm{moles}\)

5. Calculate the enthalpy change per mole of NaOH

Finally, to find the enthalpy change per mole of NaOH, divide the heat exchange (q) from step 3 by the moles of NaOH from step 4: \(\Delta H = \frac{q}{moles\,of\,NaOH}\) \(\Delta H = \frac{9.954\,\mathrm{kJ}}{0.23875\,\mathrm{moles}}\) \(\Delta H \approx 41.7\,\mathrm{kJ/mol}\) The enthalpy change for the solution process of NaOH is approximately 41.7 kJ/mol.