Question

The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by$2.54°C$ . Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is $26.42kJ/g.)$A 0.2130g sample of vanillin $\left(C_8{H}_8{O}_3\right)$is then burned in the same calorimeter, and the temperature increases by $3.25°C$. What is the energy of combustion per gram of vanillin? Per mole of vanillin?

Short answer

• Heat capacity
• Heat of combustion in
• Heat of combustion in

Step-by-step solution

Determination of heat capacity

$∆E=26.42kJ/g=4.185kJ$

$$\begin{gathered} ∆T=2.54°C \\ ∆E=heatcapacity\times ∆T \\ 4.185=heatcapacity\times 2.54°C \end{gathered}$$

$Heatcapacity=\frac{4.185kJ}{2.54°C}=1.65kJ/°C$

 Determination of heat of combustion in kJ/mol

$$\begin{gathered} ∆E=1.65kJ/°C\times 3.25=5.35kJ \\ HeatofCombustion=∆E\times n \end{gathered}$$

Where, n = moles of vanillin

$HeatofCombustion=\frac{5.35kJ}{0.2130g\times mol/152g}=3821kJ/molvanillin$

Determination of heat of combustion in kJ/g

$Heatcapacityofvanilin=\frac{∆E}{∆T}=\frac{5.35kJ}{3.25°C}=1.65kJ/°C$

Now,$Q=mc∆T=0.2130g\times 1.65kJ/°C\times 3.25°C=1.14kJ/g$