Could someone please show me how to solve this with the provided information?

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1 Answer
Aug 5, 2018

This is a Hess's law question!

Hess's law states that regardless if there are multiple steps in a reaction, the total enthalpy change for the reaction is the sum of all changes.

Explanation:

This the question you want

#C_2H_4 (g)+ H_2(g)-> C_2H_6(g)#

These are the equations given

#C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O# #H = -1411 KJ#

#C_2H_6 + 3 1/2 O_2 -> 2CO_2 + 3H_2O# #H = -1560.0 KJ#

# H_2 + 1/2 O_2 -> H_2O# #H = -285.8 KJ#

Now you need to either

Multiply the equation by a - number to flip the equation

Multiply the whole equation by a number to get more of the product or reactant.

Cancel out like reactant with like products.

Here are my steps

Here I multiplied equation 2 by a negative number to flip the
question this changes the enthalpy value from being a negative number to a positive.

I also simplified #3 1/2 O_2# to #3/2 O_2#

#C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O# #DeltaH = -1411 KJ#

# (-) 2CO_2 + 3H_2O ->C_2H_6 + 3/2 O_2# #DeltaH = +1560.0 KJ#

# H_2 + 1/2 O_2 -> H_2O# #DeltaH = -285.8 KJ#

Now you just need to cancel out like terms

#C_2H_4 + cancel 3O_2 -> cancel2CO_2 + cancel2H_2O# #DeltaH = -1411 KJ#

# (-) cancel 2CO_2 + cancel3H_2O ->C_2H_6 + cancel 3/2 O_2# #DeltaH = +1560.0 KJ#

# H_2 + cancel1/2 O_2 ->cancel H_2O# #DeltaH = -285.8 KJ#

Now your left with

#C_2H_4 (g)+ H_2(g)-> C_2H_6(g)#

Add up all your Enthalpy Values

And that's your answer

#DeltaH= -136.8 KJ#

Best Of Luck

-AN

If you have any questions please ask.