1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at

1.	1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation:C (graphite) + O2(g) → CO2(g)During the reaction, temperature rises from 298K to 299K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K. What is the enthalpy change for the above reaction at 298K and 1 atm?
Answer» Let quantity of heat from the reaction mixture = q Heat capacity of calorimeter = C_v \(\therefore\)Quantity of heat absorbed by the calorimeter q = C_v.∆T q has opposite sign because the heat is lost by the system which is equal to the heat gained by the calorimeter. \(\therefore\) q = −C_v × ∆T = −20.7kJ/K × (299 − 298)K = -20.7 kJ Here,−ve sign indicates the exothermic nature of the reaction \(\therefore\) ∆U for the combustion of the 1 g of graphite = −20.7 kJ K⁻¹ For combustion of 1 mole (12 g) of graphite = \(\frac{12g\,mol^{-1}\times(-20.7kJ)}{1g}\) = −2.48 × 10² kJ mol⁻¹ (∴ ∆n_g = 0) \(\therefore\) ∆H = ∆U = −2.48 × 10² kJ mol⁻¹

1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation:C (graphite) + O2(g) → CO2(g)During the reaction, temperature rises from 298K to 299K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K. What is the enthalpy change for the above reaction at 298K and 1 atm?

Answer»

Let quantity of heat from the reaction mixture = q

Heat capacity of calorimeter = C_v

\(\therefore\)Quantity of heat absorbed by the calorimeter

q = C_v.∆T

q has opposite sign because the heat is lost by the system which is equal to the heat gained by the calorimeter.

\(\therefore\) q = −C_v × ∆T = −20.7kJ/K × (299 − 298)K

= -20.7 kJ

Here,−ve sign indicates the exothermic nature of the reaction

\(\therefore\) ∆U for the combustion of the 1 g of graphite

= −20.7 kJ K⁻¹

For combustion of 1 mole (12 g) of graphite

= \(\frac{12g\,mol^{-1}\times(-20.7kJ)}{1g}\)

= −2.48 × 10² kJ mol⁻¹ (∴ ∆n_g = 0)

\(\therefore\) ∆H = ∆U = −2.48 × 10² kJ mol⁻¹