Jump to content
Sign in to follow this  
qwerty

Chemistry Enthalpy Reaction

Recommended Posts

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 :)

Share this post


Link to post
Share on other sites

You aren't just supposed to flip reactions and/or add them together, you also have to multiply by appropriate coeffcients. For instance you have to multiply the first reaction by 1/2 so that you get a "1" in front of C2H2.

Share this post


Link to post
Share on other sites

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?

Share this post


Link to post
Share on other sites

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.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.