Question

1 g of graphite is burnt completely in excess oxygen at 298K and 1 atmospheric pressure in a bomb calorimeter. During the reaction, the temperature raises from 298K to 299K. If the heat capacity of the bomb calorimeter is 20.7 kJ K\(^{-1}\), what is the enthalpy of combustion of C(gr)? (Atomic mass of carbon is 12 g mol\(^{-1}\))

• A. -248 kJ mol\(^{-1}\)
• B. +236 kJ mol\(^{-1}\)
• C. -236 kJ mol\(^{-1}\)
• D. +246 kJ mol\(^{-1}\)
• E. -268 kJ mol\(^{-1}\)

Hint:

Always remember that combustion is an exothermic process, so the enthalpy change (\(\Delta H\)) must be negative.
This immediately eliminates options (B) and (D).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a bomb calorimeter, the heat evolved during a reaction is measured at constant volume (\(q_v\)), which corresponds to the change in internal energy (\(\Delta U\)).
For solids like graphite, \(\Delta H\) is approximately equal to \(\Delta U\).
The heat released by the reaction is absorbed by the calorimeter.

Step 2: Key Formula or Approach:
1. Heat evolved for the given mass (\(q\)) = \(C \times \Delta T\)
2. Enthalpy of combustion (\(\Delta H_c\)) = \(\frac{q}{\text{moles of substance}} \times (-1)\) (since heat is released).

Step 3: Detailed Explanation:
Given:
Mass of Graphite = 1 g
Molar mass of Carbon = 12 g/mol
Heat capacity of calorimeter (\(C\)) = 20.7 kJ/K
Change in temperature (\(\Delta T\)) = \(299\text{ K} - 298\text{ K} = 1\text{ K}\)

Heat evolved for 1 g of graphite (\(q\)):
\[ q = C \times \Delta T = 20.7 \text{ kJ/K} \times 1 \text{ K} = 20.7 \text{ kJ} \]

Number of moles in 1 g of graphite:
\[ n = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1}{12} \text{ mol} \]

Enthalpy of combustion (per mole):
\[ \Delta H_c = -\frac{q}{n} = -\frac{20.7}{1/12} \]
\[ \Delta H_c = -20.7 \times 12 = -248.4 \text{ kJ/mol} \]
Rounding to the nearest integer gives -248 kJ/mol.

Step 4: Final Answer:
The enthalpy of combustion of C(gr) is -248 kJ mol\(^{-1}\).