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`        plz explain it..............................................................................................................................................................`
2 months ago

Vikas TU
8729 Points
```							Dear Student,Given equation areC2H4(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 asThe required equation isC2H4(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/ molor, C2H4(g)+H2(g) ---> C2H6(g)  ΔH = -559.2 kJ/ mol
```
2 months ago
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