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 textgraphite)+O2(g) → CO2(g) During the reaction, temperature rises from 298 K to 299 K If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm ?

Question

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 textgraphite)+O2(g) → CO2(g) During the reaction, temperature rises from 298 K to 299 K If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm ? (A) 20.7 kJ (B) -2.48 × 102 kJ mol-1 (C) +2.48 × 102 kJ mol-1 (D) -3.5 × 10-3 kJ mol-1

Tardigrade Admin · Accepted Answer

Heat evolved in the reaction =Cv Δ T =-20.7 × 1 = -20.7 kJ (Negative value indicates the exothermic nature of the reaction) Heat evolved in the combustion of 1 mole of graphite, i.e. 12 g graphite will be Δ U =-20.7 × 12 kJ mol-1 =-2.48 × 102 kJ mol-1 Δ H = Δ U+ΔgRT Δ ng=0 for the given reaction ∴ Δ H= Δ U =-2.48 × 102 KJ mol-1

$20.7 k J$

$- 2.48 \times 1 0^{2} k J m o l^{- 1}$

$+ 2.48 \times 1 0^{2} k J m o l^{- 1}$

$- 3.5 \times 1 0^{- 3} k J m o l^{- 1}$

20.7kJ

−2.48×102kJmol−1

+2.48×102kJmol−1

−3.5×10−3kJmol−1

$20.7 k J$

$- 2.48 \times 1 0^{2} k J m o l^{- 1}$

$+ 2.48 \times 1 0^{2} k J m o l^{- 1}$

$- 3.5 \times 1 0^{- 3} k J m o l^{- 1}$