the enthalpy of combustion of benzoic acid ($c_6h_5cooh$) is commonly used as the standard for calibrating…

the enthalpy of combustion of benzoic acid ($c_6h_5cooh$) is commonly used as the standard for calibrating constant - volume bomb calorimeters; its value has been accurately determined to be $- 3226.7\frac{kj}{mol}$. when 2.9223 g of benzoic acid are burned in a calorimeter, the temperature rises from 20.84 $^{circ}c$ to 26.67 $^{circ}c$. what is the heat capacity of the bomb? (assume that the quantity of water surrounding the bomb is exactly 2250 g.) be sure your answer has the correct number of significant digits. note: reference the phase change properties of pure substances table for additional information.

the enthalpy of combustion of benzoic acid ($c_6h_5cooh$) is commonly used as the standard for calibrating constant - volume bomb calorimeters; its value has been accurately determined to be $- 3226.7\frac{kj}{mol}$. when 2.9223 g of benzoic acid are burned in a calorimeter, the temperature rises from 20.84 $^{circ}c$ to 26.67 $^{circ}c$. what is the heat capacity of the bomb? (assume that the quantity of water surrounding the bomb is exactly 2250 g.) be sure your answer has the correct number of significant digits. note: reference the phase change properties of pure substances table for additional information.

Answer

Explanation:

Step1: Calculate moles of benzoic acid

First, find the molar - mass of benzoic acid ($C_7H_6O_2$). $M=(7\times12.01 + 6\times1.01+2\times16.00)\ g/mol=122.12\ g/mol$ $n=\frac{m}{M}=\frac{2.9223\ g}{122.12\ g/mol}=0.02393\ mol$

Step2: Calculate heat released by benzoic acid

The enthalpy of combustion of benzoic acid is $\Delta H_{comb}=- 3226.7\ kJ/mol$. $q_{rxn}=n\times\Delta H_{comb}=0.02393\ mol\times(-3226.7\ kJ/mol)= - 76.21\ kJ$ The heat absorbed by the calorimeter and water is $q_{cal}=-q_{rxn}=76.21\ kJ$

Step3: Calculate the temperature change

$\Delta T=T_2 - T_1=26.67^{\circ}C - 20.84^{\circ}C = 5.83^{\circ}C$

Step4: Calculate the heat capacity of the bomb

The heat absorbed by the calorimeter and water is given by $q_{cal}=C_{cal}\Delta T$. We know that $q_{cal}=76.21\ kJ$ and $\Delta T = 5.83^{\circ}C$. $C_{cal}=\frac{q_{cal}}{\Delta T}=\frac{76.21\ kJ}{5.83^{\circ}C}=13.1\ kJ/^{\circ}C$

Answer:

$13.1\ kJ/^{\circ}C$