0.3 g of ethane undergoes combustion at 27° C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5° C. The heat evolved during combustion of ethane at constant pressure is kJ mol -1. (Nearest integer) [Given: The heat capacity of the calorimeter system is 20 kJ K -1, R =8.3 JK -1 mol -1. Assume ideal gas behaviour. Atomic mass of C and H are 12 and 1 g mol -1 respectively]

Question

0.3 g of ethane undergoes combustion at 27° C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5° C. The heat evolved during combustion of ethane at constant pressure is kJ mol -1. (Nearest integer) [Given: The heat capacity of the calorimeter system is 20 kJ K -1, R =8.3 JK -1 mol -1. Assume ideal gas behaviour. Atomic mass of C and H are 12 and 1 g mol -1 respectively] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(Bomb calorimeter arrow const volume Heat released By combustion of 1 mole C 2 H 6(Δ U )=-(20 × 0.5/0.3) × 30=-1000 kJ C 2 H 6( g )+7 / 2 O 2( g ) arrow 2 CO 2( g )+3 H 2 O (l) Δ ng =2-(2+7 / 2)=-(7 / 2) Δ H =Δ U +Δ nRT =-1000-7 / 2 × 8.3 × 300 k J =-1000-6.225 =-1006 kJ So heat released =1006 kJ mol -1

(Bomb calorimeter → const volume Heat released By combustion of 1 mole C2​H6​(ΔU)=−0.320×0.5​×30=−1000kJ C2​H6​(g)+7/2O2​(g)→2CO2​(g)+3H2​O(l) Δng=2−(2+7/2)=−(7/2) ΔH=ΔU+ΔnRT =−1000−7/2×8.3×300kJ =−1000−6.225 =−1006kJ So heat released =1006kJmol−1