Question

How do I find `K` and `DeltaH` for the overall reaction? The overall reaction is: `"Ni"("CO")_4(g) + 4"H"_2"O"(g) rightleftharpoons "Ni"(s) + 4"CO"_2(g) + 4"H"_2(g)`

How do I find `K` and `DeltaH` for the overall reaction?

`"Ni"("CO")_4(g) rightleftharpoons "Ni"(s) + 4"CO"(g)`, `K_1 = 1.1^A`, `DeltaH_1 = 4D`
`"MnO"(s) + "CO"(g) rightleftharpoons "Mn"(s) + "CO"_2(g)`, `K_2 = B^(2.7)`, `DeltaH_2 = 340/E`
`"Mn"(s) + "H"_2"O"(g) rightleftharpoons "MnO"(s) + "H"_2(g)`, `K_3 = 1.84^C`, `DeltaH_3 = -5.77F`

The overall reaction is:
`"Ni"("CO")_4(g) + 4"H"_2"O"(g) rightleftharpoons "Ni"(s) + 4"CO"_2(g) + 4"H"_2(g)`

`A = 2`
`B = 3`
`C = 4`
`D = 5`
`E = 6`
`F = 7`

Answer 1

It really makes the most sense to identify which species are intermediates and catalysts first... That's how anyone can blindly achieve the overall reaction, simply by trying to cancel out intermediates and catalysts, or even just trying to get the products to match up.

INTERMEDIATES + CATALYSTS

Intermediates are as they seem to be; they appear in the middle of the mechanism, and go away after the reaction is over.

Catalysts come in to speed up the reaction, but are not consumed by the time the reaction is over.

By inspection, you should find:

• `"Mn"(s)` is an intermediate; appears in step 2, consumed in step 3.
• `"CO"(g)` is an intermediate. Appears in step 1, must be consumed in step 2 in order to not be in the overall reaction.
• `"MnO"(s)` is a catalyst. Consumed in step 2, reformed in step 3.

So, all you have to do is multiply each reaction step's coefficient by a constant in order to cancel out the intermediates and catalyst. This isn't nearly as complicated as it seems, because no steps need to be reversed.

FORMING OVERALL REACTION

Since you want `4"H"_2(g)` as a product, multiply step 3 by `4`. Since you want `"CO"(g)` to go away, multiply step 2 by `4` so that `4"CO"(g)` is a reactant in step 2 that gets used up from step 1.

Then by magic, the rest works itself out...

`"Ni"("CO")_4(g) rightleftharpoons "Ni"(s) + cancel(4"CO"(g))`

`4(cancel("MnO"(s)) + cancel("CO"(g)) rightleftharpoons cancel("Mn"(s)) + "CO"_2(g))`

`4(cancel("Mn"(s)) + "H"_2"O"(g) rightleftharpoons cancel("MnO"(s)) + "H"_2(g))`

`"----------------------------------------------------------------"`

`"Ni"("CO")_4(g) + 4"H"_2"O"(g) rightleftharpoons "Ni"(s) + 4"CO"_2(g) + 4"H"_2(g)`

EQUILIBRIUM CONSTANTS AND ENTHALPIES

Now, it's just a matter of recalling the general rules of `K_(eq)` and `DeltaH` with regards to scaling.

• For all thermodynamic functions, scaling a reaction multiplies the function by the constant.
• For all equilibrium constants, scaling the reaction by a constant multiplies all the exponents by that constant (since equilibria are elementary reactions), i.e. raise `bb(K_(eq))` to that constant.

Thus...

`K_1' = K_1` `" "" "` `DeltaH_1' = DeltaH_1`
`K_2' = (K_2)^4` `" "` `DeltaH_2' = 4DeltaH_2`
`K_3' = (K_3)^4` `" "` `DeltaH_3' = 4DeltaH_3`

And so... noting that adding sequential steps multiplies equilibrium constants and adds thermodynamic functions for each step by Hess's Law...

`color(blue)(K_(t ot)') = K_1'K_2'K_3' = K_1(K_2K_3)^4`

`= 1.1^A cdot (B^2.7 1.84^C)^4`

`= color(blue)(1.1^A B^10.8 1.84^(4C))`

And...

`color(blue)(DeltaH_(t ot)') = DeltaH_1' + DeltaH_2' + DeltaH_3'`

`= DeltaH_1 + 4(DeltaH_2 + DeltaH_3)`

`= 4D + 4(340/E - 5.77F)`

`= color(blue)(4(D + 340/E - 5.77F) " kJ")`

You're given the constants at the top of the page. So, plug them in and you can get your answer.