Question

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

Answer 1

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.