Question

The heat capacity of a bomb calorimeter was determined by burning 6.79 g of methane$\left(energyofcombustion=-802kJ/molC{H}_4\right)$ in the bomb. The temperature changed by $10.8^{0}C$.

a. What is the heat capacity of the bomb?

b. A 12.6g sample of acetylene$\left(C_2{H}_2\right)$ produced a temperature increase of$16.9^{0}C$ in the same calorimeter. What is the energy of combustion of acetylene $\left(inkJ/mol\right)$?

Short answer

a. Heat capacity of the bomb is $31.5kJ{/}^{0}C$

b. The energy of combustion of acetylene is $1.10x103Kj/mol$-

Step-by-step solution

The heat capacity (C) of the bomb is heatabsorbedincreaseintemprature

The energy released by the reaction = energy absorbed by the solution

From given values:Mass of methane = 6.79gm

Moles of methane =$\frac{givenmass}{molecularmass}$ =$\frac{6.79g}{16.00g/mol}=0.0424mol$

Since, energy of combustion of methane = -802 kJ/mol (given)

Therefore, heat release during the combustion of 0.0424mol of methane

$$\begin{gathered} =0.0424mol\times 802kJ/mol \\ =340kJ \end{gathered}$$

Since, the temperature changed = 10.8⁰C.

Therefore, the heat capacity ‘C’ of the bomb =

$\frac{heatabsorbed\left(q\right)}{tempraturedifference\left(\Delta T\right)}=\frac{340kj}{10.8^{0}C}=31.5kj{/}^{0}C$

to calculate the energy of combustion of acetylene C2H2:

We first find out the heat loss by acetylene given values of mass and change in temperature as follows:

Since, heat loss by the acetylene $\left(C_2{H}_2\right)$= heat gain by the calorimeter

Heat loss by $\left(C_2{H}_2\right)=$increase in temperature x heat capacity of bomb calorimeter $$\begin{gathered} =16.90C\times 31.5kJ/0C \\ =532kJ \end{gathered}$$

From the formula of the heat of combustion,

$$\begin{gathered} {\left(∆E\right)}_{comb}=heatloss\times \frac{molecularmass}{givenmass} \\ =-532kJ\times \frac{26.04g}{12.6g\left(C_2{H}_2\right)} \\ =-1.10x{10}^{3}kJ/mol \end{gathered}$$

Hence the heat of combustion of acetylene $\left(C_2{H}_2\right)=-1.10x{10}^{3}Kj/mol$