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A 6.55 g sample of aniline [tex]$\left( C_6H_5NH_2 \right.$[/tex], molar mass [tex]$\left. 93.13 \, g/mol \right)$[/tex] was combusted in a bomb calorimeter. If the temperature rose by [tex]$32.9^{\circ} C$[/tex], use the information below to determine the heat capacity of the calorimeter.

[tex]\[
4 C_6H_5NH_2(t) + 35 O_2(g) \rightarrow 24 CO_2(g) + 14 H_2O(g) + 4 NO_2(g)
\][/tex]
[tex]\[
\Delta H^{\circ}_{\text{rxn}} = -1.28 \times 10^3 \, kJ
\][/tex]

Choose the correct heat capacity:

A. [tex]$97.3 \, kJ / {^{\circ} C}$[/tex]

B. [tex]$38.9 \, kJ / {^{\circ} C}$[/tex]

C. [tex]$5.94 \, kJ / {^{\circ} C}$[/tex]

D. [tex]$12.8 \, kJ / {^{\circ} C}$[/tex]

E. [tex]$6.84 \, kJ / {^{\circ} C}$[/tex]



Answer :

GinnyAnswer
To determine the heat capacity of the calorimeter, we will go through the following steps:

1. Calculate the number of moles of aniline combusted:
Given the mass of the aniline sample is 6.55 grams and the molar mass of aniline is 93.13 g/mol, we use the formula:
[tex]\[ \text{moles of aniline} = \frac{\text{mass of sample}}{\text{molar mass}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of aniline} = \frac{6.55 \text{ g}}{93.13 \text{ g/mol}} = 0.07033 \text{ moles} \][/tex]

2. Determine the enthalpy change for the combustion of the sample:
From the reaction:
[tex]\[ 4 C_6H_5NH_2 (t) + 35 O_2(g) \rightarrow 24 CO_2(g) + 14 H_2O(g) + 4 NO_2(g) \][/tex]
the enthalpy change [tex]\(\Delta H_{rxn}\)[/tex] for the reaction is [tex]\(-1.28 \times 10^3 \text{ kJ}\)[/tex] for 4 moles of aniline.

We need to find the enthalpy change for the combustion of 0.07033 moles of aniline. Since the given [tex]\(\Delta H_{rxn}\)[/tex] is for 4 moles, the heat released (enthalpy change) for our amount is:
[tex]\[ \text{heat released} = \text{moles of aniline} \times \frac{\Delta H_{rxn}}{4} \][/tex]
Plugging in the values:
[tex]\[ \text{heat released} = 0.07033 \text{ moles} \times \frac{-1.28 \times 10^3 \text{ kJ}}{4} = -22.506 \text{ kJ} \][/tex]
The negative sign indicates heat is released (exothermic reaction).

3. Calculate the heat capacity of the calorimeter:
Using the equation for heat capacity [tex]\(C\)[/tex]:
[tex]\[ Q = C \cdot \Delta T \][/tex]
where [tex]\(Q\)[/tex] is the heat released and [tex]\(\Delta T\)[/tex] is the temperature change. Rearranging for [tex]\(C\)[/tex]:
[tex]\[ C = \frac{Q}{\Delta T} \][/tex]
We need to use the absolute value of the heat released because heat capacity is a positive quantity:
[tex]\[ C = \frac{22.506 \text{ kJ}}{32.9 \text{ °C}} = 0.684 \text{ kJ/°C} \][/tex]

Thus, the heat capacity of the calorimeter is approximately [tex]\(6.84 \text{ kJ/°C}\)[/tex].

Among the given options, the correct answer is:
[tex]\[ \boxed{6.84 \text{ kJ/°C}} \][/tex]

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