0.16 g of methane was subjected to combustion at 27 ° C in a bomb calorimeter system. The temperature of the calorimeter system (including water ) was found to rise by 0.5 ° C. Calculate the heat of combustion of methane at (i) constant volume, and (ii) constant pressure. The thermal capacity of the calorimeter system is 17.7 kJ K -1 ( R = 8.314 kJ K -1 mol -1 )
Answers
Answered by
85
CH4(g)+2O2→CO2(g)+2H2O(l)
internal energy is equal to the heat of combustion
i.e Ov=ΔE
ΔE=Heat capacity of calorimeter × rise in temperature × molar mass of compound/mass of compound
=17.7×0.5×16/0.16=885
O∨=ΔE=-885kjmol^-
heat of combustion at constant volume is equal to the enthalpy
i.e OP=ΔH
HERE
ΔH=ΔE+ΔnRT................(2)
Δn=1-3=-2,T=300K,R=8.314×10^-3kj k^-1mol6-1
Hence,from the equation(2)ΔH=-885+(-2)×8.314×10^-3×300=-890kjmo^-1
internal energy is equal to the heat of combustion
i.e Ov=ΔE
ΔE=Heat capacity of calorimeter × rise in temperature × molar mass of compound/mass of compound
=17.7×0.5×16/0.16=885
O∨=ΔE=-885kjmol^-
heat of combustion at constant volume is equal to the enthalpy
i.e OP=ΔH
HERE
ΔH=ΔE+ΔnRT................(2)
Δn=1-3=-2,T=300K,R=8.314×10^-3kj k^-1mol6-1
Hence,from the equation(2)ΔH=-885+(-2)×8.314×10^-3×300=-890kjmo^-1
Answered by
56
amount of heat required= o.5 degree= 17.7 x 0.5 = 8.85 KJ
0.16g of methane produce 8.85KJ
heat liberated by 1 mol of methane= 8.85/ 0.16 x 16 = 885 KJ
change in heat = change in energy + nRT
change of n= -2
ΔH= -885 + (-2) (8.31 x 10⁻³ x 300)
=-889.986 KJ/mol
0.16g of methane produce 8.85KJ
heat liberated by 1 mol of methane= 8.85/ 0.16 x 16 = 885 KJ
change in heat = change in energy + nRT
change of n= -2
ΔH= -885 + (-2) (8.31 x 10⁻³ x 300)
=-889.986 KJ/mol
Similar questions