Vikas TU
Last Activity: 5 Years ago
Dear Student,
Given equation are
C2H4(g)+3O2(g) ---> 2C02+2H20(l) ΔH =-1411KJ ............... (1)
2C2H6(g)+7O2(g) ---> 4CO2(G)+6H2O(l) ΔH =-372.8 KJ............... (2)
2H2(g)+O2(g) ---> 2H2O(l) ΔH =-68.3 KJ.................(3)
The enthalpy of the following reaction can be calculated as
The required equation is
C2H4(g)+H2(g) ---> C2H6(g) ΔH = ?
Divide eqn. (3) by 2 and add to the eqn (1)
C2H4(g)+3O2(g) +H2(g)+ 1/2O2(g) ---> 2CO2+2H20 + H2O(l)
ΔH= -(-337.2 -68.3/2) = 371.35 kJ/mol....................................(4)
Divide eqn. (2) by 2 and Subtract this eqn. from eqn. (4)
C2H4(g)+3O2(g)+H2(g)+1/2O2(g) -C2H6(g) - 7/2O2(g) -->2C02+2H20 +H2O(l) - 2CO2(G) - 3H2O(l)
ΔH = -371.35 - (-372.8/2) = -559.2 kJ/ mol
or, C2H4(g)+H2(g) ---> C2H6(g) ΔH = -559.2 kJ/ mol
