calculate the enthalpy of the reaction of  C2H4(g)+H2(g) give rise to C2H6(g)from the the following data??

  1. C2H4(g)+3O2(g) give rise to 2C02+2H20(l)   delH=-1411KJ
  2. 2C2H6(g)+7O2(g) give rise to  4CO2(G)+6H2O(l)  delH=-1560KJ
  3. 2H2(g)+O2(g) give rise to 2H2O(l)  delH=-285.8KJ

pls fast urgent I want it within 2hrs????pls plsssssssss help

Dear Student,

Given equation are

C2H4(g)+3O2(g) ---> 2C02+2H20(l)    ΔH =-1411KJ ............... (1)

2C2H6(g)+7O2(g) ---> 4CO2(G)+6H2O(l)  ΔH =-1560KJ............... (2) 

2H2(g)+O2(g) ---> 2H2O(l)    ΔH =-285.8KJ.................(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= -(1411 + 285.8/2) = -1553.9 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 = -1553.9 - (-1560/2) = -773.9 kJ/ mol

or, C2H4(g)+H2(g) ---> C2H6(g)  ΔH = -773.9 kJ/ mol

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