A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in kJ/mol) for the neutralization reaction.

• A. -25.0 kJ/mol
• B. -57.3 kJ/mol
• C. 25.0 kJ/mol
• D. 57.3 kJ/mol

Answer: (B)

Verified step by step guidance

1. Start by identifying the chemical reaction taking place. The neutralization reaction between NaOH and HNO3 can be represented as: ${\mathrm{NaOH}}_{\mathrm{(aq)}}+{\mathrm{HNO}3}_{\mathrm{(aq)}}\to {\mathrm{NaNO}3}_{\mathrm{(aq)}}+{H2O}_{\mathrm{(l)}}$.
2. Calculate the number of moles of NaOH and HNO3. Since both solutions have the same concentration and volume, the number of moles for each is: $n=c\times V$, where $c=0.300 M$ and $V=100.0 \mathrm{mL}$ (convert mL to L).
3. Determine the heat absorbed or released by the solution using the formula: $q=m\times c\times \Delta T$, where $m$ is the mass of the solution, $c$ is the specific heat capacity (assume water's specific heat capacity, $4.18 J/g\times K$), and $\Delta T$ is the change in temperature.
4. Convert the heat absorbed or released from joules to kilojoules, and then calculate the enthalpy change per mole of reaction. Use the formula: $\Delta H=qn$, where $n$ is the number of moles of the limiting reactant.
5. Consider the sign of $\Delta H$. Since the temperature increased, the reaction is exothermic, and $\Delta H$ should be negative.