Question

When 'X' g of graphite is completely burnt in a bomb calorimeter in excess of \( O_2 \) at 298 K and 1 atm pressure as given in the equation $$ \text{C(graphite)} + O_2(g) \rightarrow CO_2(g) $$ The temperature of calorimeter raised from 298 K to 302 K. If the heat capacity of the calorimeter and molar enthalpy change for the reaction at 1 atm and 298 K are 20.7 kJ K\(^{-1} \) and -248.4 kJ mol\(^{-1} \), 'X' in g is

• A. 8
• B. 2
• C. 3
• D. 4

Hint:

In a bomb calorimeter, the heat released by the reaction is absorbed by the calorimeter. Calculate the heat absorbed by the calorimeter using \( q = C \Delta T \). The heat released by the reaction is equal in magnitude. Use the molar enthalpy change to find the number of moles of the reactant burnt and then convert moles to mass using the molar mass.

Answer 1

The Correct Option is D

The heat released by the combustion of 'X' g of graphite in the bomb calorimeter causes the temperature of the calorimeter to rise.
The heat absorbed by the calorimeter \( q_{cal} \) is given by: $$ q_{cal} = C \Delta T $$ where \( C \) is the heat capacity of the calorimeter and \( \Delta T \) is the change in temperature.
Given: \( C = 20.
7 \) kJ K\(^{-1} \) Initial temperature \( T_i = 298 \) K Final temperature \( T_f = 302 \) K \( \Delta T = T_f - T_i = 302 - 298 = 4 \) K So, the heat absorbed by the calorimeter is: $$ q_{cal} = (20.
7 \text{ kJ K}^{-1}) \times (4 \text{ K}) = 82.
8 \text{ kJ} $$ Since the bomb calorimeter is an isolated system, the heat released by the reaction \( q_{rxn} \) is equal in magnitude but opposite in sign to the heat absorbed by the calorimeter: $$ q_{rxn} = -q_{cal} = -82.
8 \text{ kJ} $$ The molar enthalpy change for the reaction is given as \( \Delta H = -248.
4 \) kJ mol\(^{-1} \).
Since the reaction is carried out at constant volume in a bomb calorimeter, the heat released at constant volume \( q_v \) is approximately equal to the enthalpy change \( \Delta H \) (as the number of moles of gas is the same on both sides of the equation).
So, the heat released when 1 mole of graphite burns is 248.
4 kJ.
Let the number of moles of graphite burnt be \( n \).
The total heat released by burning 'X' g of graphite is \( n \times (-\Delta H) \).
$$ -82.
8 \text{ kJ} = n \times (-248.
4 \text{ kJ mol}^{-1}) $$ $$ n = \frac{82.
8}{248.
4} \text{ mol} = \frac{828}{2484} \text{ mol} = \frac{1}{3} \text{ mol} $$ The molar mass of graphite (Carbon) is approximately 12 g mol\(^{-1} \).
The mass 'X' of graphite burnt is given by: $$ X = n \times \text{molar mass of C} = \frac{1}{3} \text{ mol} \times 12 \text{ g mol}^{-1} = 4 \text{ g} $$ Therefore, the value of 'X' is 4 g.