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Chemistry Enthalpy Reaction


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I have been given a question:


Determine the enthalpy of reaction in kJ for the hydrogenation of acetylene to form ethane :

C2H2 (g) + 2H2(g) --> C2H6(g)


From the following data:


2C2H2 + 5O2 --> 4CO2 + 2H2O H = -2600 KJ

2C2H6 + 7O2 --> 4CO2 + 6H2O H = -3120 KJ

H2 + 1/2O2 --> H2O H = -286 KJ




So far i know H(h2o) = -286 kJ

But i've swapped around reaction 1 (the first one where H = -2600) and added together the first and second reactions.

Now i have:

2C2H6 + 2O2 ==> 2C2H2 + 4H2O H = -520 KJ


But i dont know where to go from here, because i'm left with two unknowns, the 2c2h6 and 2c2h2.


Any help appreciated. thanks guys/girls :)

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ok, but even if i did multiply the top one by .5 it would still give me the same answer in the long run.?


2C2H6 + 2O2 ==> 2C2H2 + 4H2O H = -520 KJ

The 2 unknowns i have here.. how can i figure out those?

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Okay heres how it goes:

Start with your three reactions


[math]2 C_2 H_2 + 5 O_2 -> 4 CO_2 + 2 H_2 OH = -2600 KJ[/math]


[math]2 C_2 H_6 + 7 O_2 -> 4 CO_2 + 6 H_2 OH = -3120 KJ[/math]


[math]H_2 + \frac{1}{2} O_2 ->H_2 OH = -286 KJ[/math]


Remember that you want everything to be on the correct side of the arrow when you get to the end, like:


[math]C_2 H_2 + 2 H_2 -> C_2 H_6[/math]


Step 1: Divide the first reaction by 2


[math]C_2 H_2 + \frac{5}{2} O_2 -> 2 CO_2 + H_2 OH = -1300 KJ[/math]


The second reaction reaction by 2 and switch the direction(which includes changing the sign for the energy)


[math]2 CO_2 + 3 H_2 OH-> C_2 H_6 + \frac{7}{2} O_2 = 1580 KJ[/math]


and multiply the third reaction by 2


[math]2 H_2 + O_2 -> 2 H_2 OH = -572 KJ[/math]


Now I add the three reactions together to get:


[math]C_2 H_2 + \frac{5}{2} O_2 + 2 CO_2 + 3 H_2 OH + 2 H_2 + O_2[/math] -> [math]2 CO_2 + H_2 OH + C_2 H_6 + \frac{7}{2} O_2 + 2 H_2 OH = -1300 KJ + 1580 KJ -572 KJ[/math]


It can be simplified into:


[math]C_2 H_2 + 2 H_2 ->C_2 H_6 = -292 KJ[/math]


by crossing out the compounds that are the same on both sides of the reaction.

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