Question

Given,$0.16g$ of methane was subjected to combustion at $27^\circ C$ in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by $0.5^\circ C$ . The heat of combustion of methane at
(i) Constant volume (ii) constant pressure
[Given: the thermal capacity of calorimeter system is $17.7kJ{K^{ - 1}}.\left( {R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}} \right)$]
A.$\left( i \right) - 885kJ/mol\left( {ii} \right) - 889.95kJ/mol$
B.$\left( i \right) - 785kJ/mol\left( {ii} \right) - 859.95kJ/mol$
C.$\left( i \right) - 587kJ/mol\left( {ii} \right) - 789.95kJ/mol$
D.$\left( i \right) - 985kJ/mol\left( {ii} \right) - 999.95kJ/mol$

Answer 1

We need to know that the bomb calorimeter is used for measuring the heat change in form energy in chemical and physical change. This calorimeter depends on the heat capacity. In this calorimeter heat change in the reaction is directly proportional to the heat capacity. In this calorimeter two conditions are important. One is at constant volume and at constant pressure. The symbolic representation of constant volume is ${q_v}$ and ${q_p}$.
Formula used:
Heat absorbed at constant volume,
${q_v} = {m_w}{C_w}\Delta T$
${q_p} = {q_v} + \Delta nR\Delta T$
Here ${q_v}$ is heat absorbed at constant volume
${q_p}$ is heat absorbed at pressure volume
Molar mass of the system is ${m_w}$
The gas constant is $R$
The mole difference between the product and reactant is $\Delta n$
The thermal capacity of the system is ${C_w}$
Raise in temperature at the system at degree Celsius is $\Delta T$
Conversion degree Celsius to kelvin
$\text{temperature at kelvin} = \text{temperature at degree} + 273$

Complete answer:
The combustion reaction of methane is given below,
$C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O$
The given mass of methane is $0.16g$
Molar mass of the system is $16g$
The thermal capacity of the system is $17.7kJ{K^{ - 1}}$
Temperature at the system at degree Celsius is $27^\circ C$
Temperature at the system at kelvin is $300K$
The raise in temperature is $0.5^\circ C$
The gas constant is $8.314Jmo{l^{ - 1}}{K^{_{^{ - 1}}}}$
The moles difference in gas molecule in the reaction is $(1 - 3) = - 2$
${q_v} = {m_w}{C_w}\Delta T$
Now we can substitute the known values we get,
$= 0.5 \times 17.7 \times \dfrac{{16}}{{0.16}}$
On simplification we get,
$\Rightarrow {q_v} = - 885kJ/mol$
${q_p} = {q_v} + \Delta nR\Delta T$
Now we can substitute the known values we get,
$= - 885 + ( - 2 \times 300 \times 8.314)$
On simplification we get,
${q_p} = - 889.9kJ/mol$

Hence, option A is correct, the constant volume and constant pressure of the combustion of methane is$- 885kJ/mol$ and $- 889.95kJ/mol$.

Note:
We have to know that this bomb calorimeter is used to calculate the amount of heat released in a combustion reaction in chemistry. It is also used to calculate the calories of our taken food. It is mainly used for diet patients. This bomb calorimeter is also used for metabolic study of the human body. This calorimeter is used for the explosive test of the pyrophoric compound in the laboratory.