In a coffee cup calorimeter, $14.0 \mathrm{~mL}$ of $0.65 \mathrm{M}$ nitric acid $\left(\mathrm{HNO}_{3}\right)$ and $14.0 \mathrm{~mL}$ of $0.65 \mathrm{M}$ potassium hydroxide $(\mathrm{KOH})$ are mixed to observe the heat released during the neutralization reaction. Based on the data in the table, what is the enthalpy of the neutralization reaction between $\mathrm{HNO}_{3}$ and $\mathrm{KOH}$ ?
\begin{tabular}{|c|c|}
\hline Initial temperature in the calorimeter $\left({ }^{\circ} \mathrm{C}\right)$ & 21.9 \
\hline Final temperature in the calorimeter $\left({ }^{\circ} \mathrm{C}\right)$ & 26.2 \
\hline Final mass of the neutralized solution $(\mathrm{g})$ & 27.91 \
\hline Calorimeter constant $\left.\left(\mathrm{J} /{ }^{\circ} \mathrm{C}\right)\right)$ & 4.47 \
\hline
\end{tabular}
$$
\Delta H_{\mathrm{rxn}}=\square \mathrm{kJ} / \mathrm{mol}
$$
$\square$

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the reaction**: The reaction between nitric acid ($\mathrm{HNO}_{3}$) and potassium hydroxide ($\mathrm{KOH}$) is a neutralization reaction, which can be represented as: $$\mathrm{HNO}_{3} + \mathrm{KOH} ightarrow \mathrm{KNO}_{3} + \mathrm{H}_{2}\mathrm{O}$$ ‖ ‖ 2 ⋮ **Calculate the heat absorbed by the solution**: The heat absorbed by the solution ($q_{ ext{solution}}$) can be calculated using the formula: $$q_{ ext{solution}} = m \cdot c \cdot \Delta T$$ where $m$ is the mass of the solution, $c$ is the specific heat capacity of water (4.18 J/g°C), and $\Delta T$ is the change in temperature. Given: - $m = 27.91 \, ext{g}$ - $\Delta T = 26.2 - 21.9 = 4.3 \, ext{°C}$ $$q_{ ext{solution}} = 27.91 \, ext{g} \cdot 4.18 \, ext{J/g°C} \cdot 4.3 \, ext{°C} = 501.6 \, ext{J}$$ ‖ ‖ 3 ⋮ **Calculate the heat absorbed by the calorimeter**: The heat absorbed by the calorimeter ($q_{ ext{calorimeter}}$) can be calculated using the formula: $$q_{ ext{calorimeter}} = C_{ ext{cal}} \cdot \Delta T$$ where $C_{ ext{cal}}$ is the calorimeter constant. Given: - $C_{ ext{cal}} = 4.47 \, ext{J/°C}$ - $\Delta T = 4.3 \, ext{°C}$ $$q_{ ext{calorimeter}} = 4.47 \, ext{J/°C} \cdot 4.3 \, ext{°C} = 19.2 \, ext{J}$$ ‖ ‖ 4 ⋮ **Calculate the total heat released**: The total heat released ($q_{ ext{total}}$) is the sum of the heat absorbed by the solution and the calorimeter. $$q_{ ext{total}} = q_{ ext{solution}} + q_{ ext{calorimeter}} = 501.6 \, ext{J} + 19.2 \, ext{J} = 520.8 \, ext{J}$$ ‖ ‖ 5 ⋮ **Calculate the moles of reactants**: The moles of $\mathrm{HNO}_{3}$ and $\mathrm{KOH}$ can be calculated using the molarity and volume. Given: - Volume of $\mathrm{HNO}_{3}$ and $\mathrm{KOH}$ = 14.0 mL each - Molarity of $\mathrm{HNO}_{3}$ and $\mathrm{KOH}$ = 0.65 M $$ ext{Moles of } \mathrm{HNO}_{3} = ext{Moles of } \mathrm{KOH} = 0.65 \, ext{M} imes 0.014 \, ext{L} = 0.0091 \, ext{mol}$$ ‖ ‖ 6 ⋮ **Calculate the enthalpy change per mole**: The enthalpy change of the reaction ($\Delta H_{ ext{rxn}}$) can be calculated using the formula: $$\Delta H_{ ext{rxn}} = \frac{q_{ ext{total}}}{ ext{moles of reactants}}$$ Given: - $q_{ ext{total}} = 520.8 \, ext{J} = 0.5208 \, ext{kJ}$ - Moles of reactants = 0.0091 mol $$\Delta H_{ ext{rxn}} = \frac{0.5208 \, ext{kJ}}{0.0091 \, ext{mol}} = 57.2 \, ext{kJ/mol}$$ ‖ ∻Answer∻ ⚹ The enthalpy of the neutralization reaction between $\mathrm{HNO}_{3}$ and $\mathrm{KOH}$ is $57.2 \, ext{kJ/mol}$. ⚹ ∻Key Concept∻ ⚹ Enthalpy change of a reaction can be determined using calorimetry data and the formula $\Delta H_{ ext{rxn}} = \frac{q_{ ext{total}}}{ ext{moles of reactants}}$. ⚹ ∻Explanation∻ ⚹ By calculating the heat absorbed by the solution and the calorimeter, and then dividing by the moles of reactants, we can determine the enthalpy change of the reaction. ⚹◊What is the enthalpy of the neutralization reaction between $\mathrm{HNO}_3$ and $\mathrm{KOH}$ based on the given information?⍭ Generate me a similar question◊

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