Question

Four grams of graphite is burnt in a bomb calorimeter of heat capacity $\text{ 30 kJ }{{\text{K}}^{-1}}\text{ }$in excess of oxygen at 1 atmospheric pressure. The temperature rises from $\text{ 300 K }$to$\text{ 304 K }$. What is the enthalpy of combustion of graphite (in$\text{ kJ mo}{{\text{l}}^{-1}}$ )?
A) $\text{ }360\text{ }$
B) $\text{ 1440 }$
C) $-360\text{ }$
D) $-1440\text{ }$

Answer 1

Hint: Suppose a system containing the compound A is subjected to the combustion in the calorimetry combustion, then the enthalpy of combustion i.e.
 $\text{ H = C }\times \text{ }\theta \times \dfrac{M}{m}\text{ }$
Here, H is the constant volume heat of combustion or the enthalpy of combustion, C is the thermal heat capacity of the calorimeter, $\text{ }\theta \text{ }$ is the temperature change, M is the molar mass of the substance and m is the given weight of the substance.

Complete answer:
Suppose a system containing the compound A is subjected to the combustion in the calorimetry combustion, then the enthalpy of combustion i.e.
 $\text{ H = C }\times \text{ }\theta \times \dfrac{M}{m}\text{ }$
Here, H is the constant volume heat of combustion or the enthalpy of combustion, C is the thermal heat capacity of the calorimeter, $\text{ }\theta \text{ }$ is the temperature change, M is the molar mass of the substance and m is the given weight of the substance.
We have given the following data:
Amount of graphite burnt is, $\text{ w = 4 g }$
The heat capacity of the calorimeter is given as $\text{ 30 kJ }{{\text{K}}^{-1}}\text{ }$
The pressure is equal to 1 atm
The temperature rises from $\text{ }{{\text{T}}_{\text{1}}}\text{ = 300 K }$ to ${{\text{T}}_{2}}\text{ = 304 K}$.
We are interested to find out the enthalpy of combustion of graphite in calorimetry.
Let's substitute the values in the equation, we have,
$\begin{align}
  & \text{ H = C }\times \text{ }\theta \times \dfrac{M}{m}\text{ } \\
 & \Rightarrow \left( -30\text{ kJ mo}{{\text{l}}^{-1}} \right)\times \left( 304-300 \right)\times \dfrac{12\text{ }}{4\text{ }} \\
 & \Rightarrow \left( -30\text{ kJ mo}{{\text{l}}^{-1}} \right)\times 4\times \dfrac{12\text{ }}{4\text{ }}\text{ } \\
 & \therefore \text{ H}=\text{ }-360\text{ kJ mo}{{\text{l}}^{-1}} \\
\end{align}$
Thus, heat or enthalpy of combustion for the graphite in excess of oxygen is equal to $\text{ }-360\text{ kJ mo}{{\text{l}}^{-1}}\text{ }$ .

Hence, (C) is the correct option.

Note:
It may be noted that the enthalpy of a reaction can also be written as follows,
$\text{ }{{\text{q}}_{\text{P}}}\text{ = H + }\Delta {{\text{n}}_{\text{g}}}\text{RT }$
Where ${{\text{q}}_{\text{P}}}$ is the total enthalpy of reaction, $\text{ }\Delta {{\text{n}}_{\text{g}}}\text{ }$is the difference between the number of moles of the gaseous product and gaseous reactant and T is the temperature ($\text{ 298K }$ ). $\text{ C + }{{\text{O}}_{\text{2}}}\to \text{C}{{\text{O}}_{\text{2}}}\text{ }$
$\text{ }\Delta {{\text{n}}_{\text{g}}}\text{ }$will be equal to, $\text{ 1}-(2)\text{ = }-1\text{ }$. Thus, the enthalpy of the graphite would be,
$\begin{align}
  & \text{ }{{\text{q}}_{\text{P}}}\text{ = }-360\text{ kJ mo}{{\text{l}}^{-1}}\text{ + (}-1)\left( 8.314\times {{10}^{-3}} \right)\left( 298\text{K} \right)\text{ } \\
 & \Rightarrow \text{ (}-360\text{ }-2.477)\text{kJ mo}{{\text{l}}^{-1}}\text{ } \\
 & \therefore {{\text{q}}_{\text{P}}}\text{= 362}\text{.47 kJ mo}{{\text{l}}^{-1}} \\
\end{align}$