qwerty Posted April 3, 2006 Share Posted April 3, 2006 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 Link to comment Share on other sites More sharing options...

Tom Mattson Posted April 3, 2006 Share Posted April 3, 2006 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 C_{2}H_{2}. Link to comment Share on other sites More sharing options...

qwerty Posted April 10, 2006 Author Share Posted April 10, 2006 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? Link to comment Share on other sites More sharing options...

s pepperchin Posted April 10, 2006 Share Posted April 10, 2006 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. Link to comment Share on other sites More sharing options...

## Recommended Posts

## 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