To determine the heat capacity of the calorimeter in this problem, let's follow a step-by-step approach:
1. Calculate the moles of aniline combusted:
- We start with the mass of the aniline sample, which is 6.55 grams.
- The molar mass of aniline is given as 93.13 g/mol.
- Using the formula for moles ([tex]\( n \)[/tex]), which is [tex]\(\frac{\text{mass}}{\text{molar mass}}\)[/tex], we get:
[tex]\[
n = \frac{6.55 \text{ g}}{93.13 \text{ g/mol}} = 0.07033179426607968 \text{ mol}
\][/tex]
2. Calculate the total heat released during the combustion:
- The enthalpy change of the reaction ([tex]\(\Delta H_{rxn}\)[/tex]) is given as [tex]\( -1.28 \times 184 \text{ kJ/mol} \)[/tex].
- First, compute [tex]\(\Delta H_{rxn}\)[/tex]:
[tex]\[
\Delta H_{rxn} = -1.28 \times 184 = -235.52 \text{ kJ/mol}
\][/tex]
- Then, multiply the number of moles of aniline by the enthalpy change to find the heat released ([tex]\( q \)[/tex]):
[tex]\[
q = 0.07033179426607968 \text{ mol} \times (-235.52 \text{ kJ/mol}) = -16.564544185547085 \text{ kJ}
\][/tex]
3. Determine the heat capacity of the calorimeter:
- The temperature rise observed is 32.9 [tex]\( ^\circ \text{C} \)[/tex].
- Heat capacity ([tex]\( C_{cal} \)[/tex]) is defined by the equation:
[tex]\[
C_{cal} = \frac{-q}{\Delta T}
\][/tex]
where [tex]\( q \)[/tex] is the heat released and [tex]\( \Delta T \)[/tex] is the temperature rise.
- Using the values we have:
[tex]\[
C_{cal} = \frac{16.564544185547085 \text{ kJ}}{32.9 ^\circ \text{C}} = 0.5034815861868415 \text{ kJ/} ^\circ \text{C}
\][/tex]
4. Match the calculated heat capacity to the closest provided option:
- The options given are 97.3 kJ/[tex]\( ^\circ \text{C} \)[/tex], 38.9 kJ/[tex]\( ^\circ \text{C} \)[/tex], 5.94 kJ/[tex]\( ^\circ \text{C} \)[/tex], 12.8 kJ/[tex]\( ^\circ \text{C} \)[/tex], and 6.84 kJ/[tex]\( ^\circ \text{C} \)[/tex].
- Compare each option to the calculated value of 0.5034815861868415 kJ/[tex]\( ^\circ \text{C} \)[/tex]:
The closest option is 5.94 kJ/[tex]\( ^\circ \text{C} \)[/tex].
5. Conclusion:
- Therefore, the heat capacity of the calorimeter is closest to 5.94 kJ/° C.
So, the heat capacity of the calorimeter is [tex]\(\boxed{5.94 \text{ kJ/}^\circ \text{C}}\)[/tex].
