Question

When \(2.0 \mathrm{g}\) of \(\mathrm{NaOH}\) were dissolved in \(53.0 \mathrm{g}\) water in a calorimeter at \(24.0^{\circ} \mathrm{C}\), the temperature of the solution went up to \(33.7^{\circ} \mathrm{C}\). a. Is this solution reaction exothermic? ______ Why? b. Calculate \(q_{H_{2} o},\) using Eq. 1 _______ joules c. Find \(\Delta H\) for the reaction as it occurred in the calorimeter (Eq. 5 ). \(\Delta H=\) ________ joules d. Find \(\Delta H\) for the solution of \(1.00 \mathrm{g} \mathrm{NaOH}\) in water. \(\Delta H=\) _______ joules/g e. Find \(\Delta H\) for the solution of one mole \(\mathrm{NaOH}\) in water. \(\Delta H=\) _______ joules/mole f. Given that NaOH exists as \(\mathrm{Na}^{+}\) and \(\mathrm{OH}^{-}\) ions in solution, write the equation for the reaction that occurs when \(\mathrm{NaOH}\) is dissolved in water. g. Using cnthalpies of formation as given in thermodynamic tables, calculate \(\Delta H\) for the reaction in Part f and compare your answer with the result you obtained in Part e.

Short answer

The reaction is exothermic as the temperature increased from \(24.0^{\circ} \mathrm{C}\) to \(33.7^{\circ} \mathrm{C}\). The heat absorbed by the water, \(q_{H_{2}o}\), is found to be \(2183.14 \mathrm{J}\). The enthalpy change, \(\Delta H\), is \(-2183.14 \mathrm{J}\) for the reaction in the calorimeter, \(-1091.57 \mathrm{J/g}\) for the solution of \(1.00 \mathrm{g} \mathrm{NaOH}\) in water, and \(-43659.82 \mathrm{J/mol}\) for the solution of one mole NaOH in water. The reaction when NaOH is dissolved in water is: \(NaOH(s) \rightarrow Na^{+}(aq) + OH^{-}(aq)\). Using enthalpies of formation, the \(\Delta H\) calculated is \(-750.00 \mathrm{J/mol}\), which is different from the value obtained in Part e due to experimental factors.

Step-by-step solution

a. Determine if the solution is exothermic

To see if the reaction is exothermic or not, we will look at the initial and final temperature of the solution. If the temperature increased, then heat was released, and the reaction is exothermic. In this case, the initial temperature was \(24.0^{\circ} \mathrm{C}\) and the final temperature was \(33.7^{\circ} \mathrm{C}\). Since the final temperature is higher, this reaction is exothermic.

b. Calculate \(q_{H_{2} o}\) using Eq. 1

We will use the equation: \(q=m×c×\Delta T\), where \(q\) is the heat absorbed/released, \(m\) is the mass of the substance, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. For water, the specific heat capacity \(c=4.18 \mathrm{J/g^{\circ} C}\). Let's calculate the heat absorbed/released by water: \(q_{H_{2} o} = (53.0 \mathrm{g})(4.18 \mathrm{J/g^{\circ} C})(33.7^{\circ} \mathrm{C} - 24.0^{\circ} \mathrm{C})\) \(q_{H_{2} o} = 2183.14 \mathrm{J}\)

c. Calculate \(\Delta H\) for the reaction as it occurred in the calorimeter

Since the reaction is exothermic and heat is released in the system, the heat absorbed by the water has the same magnitude but opposite sign as the heat released by the reaction: \(\Delta H = -q_{H_{2} o} = -2183.14 \mathrm{J}\)

d. Calculate \(\Delta H\) for the solution of 1.00 g NaOH in water

To find the enthalpy change for dissolving 1.00 g NaOH, we will divide the enthalpy calculated in step c by the initial mass of NaOH (2.00 g). \(\Delta H = \frac{-2183.14 \mathrm{J}}{2.00 \mathrm{g}} = -1091.57 \mathrm{J/g}\)

e. Calculate \(\Delta H\) for the solution of one mole NaOH in water

Using the molar mass of NaOH (22.99 g/mol Na + 15.999 g/mol O + 1.0078 g/mol H = 39.996 g/mol), we will multiply the enthalpy change for 1 g NaOH by the molar mass to find the enthalpy change for one mole of NaOH: \(\Delta H = (-1091.57 \mathrm{J/g})(39.996 \mathrm{g/mol}) = -43659.82 \mathrm{J/mol}\)

f. Write the equation for the reaction when NaOH is dissolved in water

When NaOH is dissolved in water, it dissociates into Na+ and OH- ions: \(NaOH(s) \rightarrow Na^{+}(aq) + OH^{-}(aq)\)

g. Calculate \(\Delta H\) using enthalpies of formation and compare with Part e

Using the enthalpies of formation from thermodynamic tables, we have: \(\Delta H = \Delta H_{f}(Na^{+}) + \Delta H_{f}(OH^{-}) - \Delta H_{f}(NaOH)\) According to the tables, \(\Delta H_{f}(Na^{+}) = -240.1\mathrm{kJ/mol}\), \(\Delta H_{f}(OH^{-}) = -229.9\mathrm{kJ/mol}\), and \(\Delta H_{f}(NaOH) = -469.25 \mathrm{kJ/mol}\). Therefore, \(\Delta H = -240.1 - 229.9 + 469.25 = -0.75\mathrm{kJ/mol} = -750.00 \mathrm{J/mol}\) Comparing this value to the result from Part e (\(-43659.82 \mathrm{J/mol}\)), we see there is a difference in values. The difference can be attributed to heat losses during the experiment, the assumption that the specific heat capacity of the solution is equal to water, and other experimental factors.

Key concepts

Exothermic Reaction

An exothermic reaction is a chemical reaction that releases energy in the form of heat or light. This is the opposite of an endothermic reaction, which absorbs energy. In an exothermic process, the energy of the reactants is higher than that of the products, resulting in the excess energy being released to the surroundings.

One clear indicator that a reaction is exothermic is an increase in temperature of the solution when the reaction occurs. For instance, in the exercise, when sodium hydroxide (\(\mathrm{NaOH}\)) is dissolved in water, the solution's temperature increased from \(24.0^{\circ}\mathrm{C}\) to \(33.7^{\circ}\mathrm{C}\). This visible temperature rise implies that the reaction released heat into the surroundings.

• Energy is released (surroundings get warmer).
• Temperature increase signifies an exothermic reaction.

Understanding exothermic reactions is crucial in calorimetry, as they help us determine the amount of energy change involved in reactions.

Enthalpy Change

The enthalpy change (\(\Delta H\)) is a measurement of the total heat content in a system. It reflects the amount of energy absorbed or released during a reaction at constant pressure. For exothermic reactions, the enthalpy change is negative because energy is released.

In the original exercise, the enthalpy change is calculated by measuring the heat absorbed by the water, using the formula \(q=m \times c \times \Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature difference.

• A negative \(\Delta H\) indicates an exothermic reaction.
• Calculated based on heat transfer to the surroundings.

The magnitude of \(\Delta H\) can vary depending on whether it is per gram or per mole of a substance. As shown in the step-by-step solution, \(\Delta H\) was calculated per gram, and then adjusted to represent the change per mole. This is essential for understanding the energy changes related to different quantities of reactants.

Thermochemical Equations

Thermochemical equations are balanced chemical equations that include both the quantity of products and reactants and the enthalpy change. These equations help convey the energy change happening during the reaction.

In the exercise, when \(\mathrm{NaOH}\) is dissolved in water, the reaction can be represented as:
\[ \mathrm{NaOH(s) \rightarrow Na^{+}(aq) + OH^{-}(aq)} \]
This equation shows that the solid sodium hydroxide dissociates into sodium and hydroxide ions in the solution. It is also crucial to consider the enthalpy values usually included in these equations, indicating the energy change.

• Shows both reactants/products and energy change.
• Used for calculations and energy tracking.

Comparing calculated \(\Delta H\) from experiments and thermochemical equations is crucial for validating experimental results and understanding discrepancies, such as heat loss in a calorimeter setup. This comparison also aids in discussing potential sources of error in heat capacity assumptions and measurement techniques.