Questions: In an experiment, a 0.3208 g sample of biphenyl (C12H10) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.191 x 10^3 g of water. During the combustion the temperature increases from 27.33 to 29.65°C. The heat capacity of water is 4.184 J · g^-1 · °C^-1. The heat capacity of the calorimeter was determined in a previous experiment to be 795.1 J · °C^-1. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of biphenyl based on these data. C12H10(s)+(29 / 2) O2(g) -> 5 H2O(l)+12 CO2(g)+ Energy Molar Heat of Combustion = kJ / mol

In an experiment, a 0.3208 g sample of biphenyl (C12H10) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.191 x 10^3 g of water. During the combustion the temperature increases from 27.33 to 29.65°C. The heat capacity of water is 4.184 J · g^-1 · °C^-1.
The heat capacity of the calorimeter was determined in a previous experiment to be 795.1 J · °C^-1.
Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of biphenyl based on these data.
C12H10(s)+(29 / 2) O2(g) -> 5 H2O(l)+12 CO2(g)+ Energy

Molar Heat of Combustion = kJ / mol

Transcript text: In an experiment, a 0.3208 g sample of biphenyl $\left(\mathrm{C}_{12} \mathrm{H}_{10}\right)$ is burned completely in a bomb calorimeter. The calorimeter is surrounded by $1.191 \times 10^{3} \mathrm{~g}$ of water. During the combustion the temperature increases from 27.33 to $29.65^{\circ} \mathrm{C}$. The heat capacity of water is $4.184 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}$. The heat capacity of the calorimeter was determined in a previous experiment to be $795.1 \mathrm{~J} \cdot{ }^{\circ} \mathrm{C}^{-1}$. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of biphenyl based on these data. \[ \mathrm{C}_{12} \mathrm{H}_{10}(s)+(29 / 2) \mathrm{O}_{2}(g) \rightarrow 5 \mathrm{H}_{2} \mathrm{O}(l)+12 \mathrm{CO}_{2}(g)+\text { Energy } \] Molar Heat of Combustion = $\square$ $\mathrm{kJ} / \mathrm{mol}$

Solution

Solution Steps

Step 1: Calculate the Temperature Change

Initial temperature: $27.33^{\circ} \mathrm{C}$
Final temperature: $29.65^{\circ} \mathrm{C}$
Temperature change ($\Delta T$): \[ \Delta T = 29.65 - 27.33 = 2.32^{\circ} \mathrm{C} \]

Step 2: Calculate the Heat Absorbed by Water

Mass of water: $1.191 \times 10^{3} \mathrm{~g}$
Specific heat capacity of water: $4.184 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}$
Heat absorbed by water ($q_{\text{water}}$): \[ q_{\text{water}} = \text{mass} \times \text{specific heat capacity} \times \Delta T \] \[ q_{\text{water}} = 1.191 \times 10^{3} \mathrm{~g} \times 4.184 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \times 2.32^{\circ} \mathrm{C} \] \[ q_{\text{water}} = 11577.5 \mathrm{~J} \]

Step 3: Calculate the Heat Absorbed by the Calorimeter

Heat capacity of the calorimeter: $795.1 \mathrm{~J} \cdot{ }^{\circ} \mathrm{C}^{-1}$
Heat absorbed by the calorimeter ($q_{\text{calorimeter}}$): \[ q_{\text{calorimeter}} = \text{heat capacity} \times \Delta T \] \[ q_{\text{calorimeter}} = 795.1 \mathrm{~J} \cdot{ }^{\circ} \mathrm{C}^{-1} \times 2.32^{\circ} \mathrm{C} \] \[ q_{\text{calorimeter}} = 1844.6 \mathrm{~J} \]

Step 4: Calculate the Total Heat Released

Total heat released ($q_{\text{total}}$): \[ q_{\text{total}} = q_{\text{water}} + q_{\text{calorimeter}} \] \[ q_{\text{total}} = 11577.5 \mathrm{~J} + 1844.6 \mathrm{~J} \] \[ q_{\text{total}} = 13422.1 \mathrm{~J} \]

Step 5: Calculate the Molar Heat of Combustion

Molar mass of biphenyl ($\mathrm{C}_{12} \mathrm{H}_{10}$): \[ \text{Molar mass} = 12 \times 12.01 + 10 \times 1.008 = 154.18 \mathrm{~g/mol} \]
Moles of biphenyl burned: \[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.3208 \mathrm{~g}}{154.18 \mathrm{~g/mol}} = 0.00208 \mathrm{~mol} \]
Molar heat of combustion ($\Delta H_{\text{combustion}}$): \[ \Delta H_{\text{combustion}} = \frac{q_{\text{total}}}{\text{moles}} = \frac{13422.1 \mathrm{~J}}{0.00208 \mathrm{~mol}} \] \[ \Delta H_{\text{combustion}} = 64529.3 \mathrm{~J/mol} = 64.53 \mathrm{kJ/mol} \]

Step 6: Final Answer