Determine Kp for N2(g) + 3H2 (g) \Updownarrow 2NH3(g)

[312213]

Homework Statement

H₂ (g) + Cl₂(g) $$\Updownarrow$$ 2HCl (g) K_p = 2.5 × 10³³
NH₃(g) + HCl(g) $$\Updownarrow$$ NH₄Cl(s) K_p = 2.1 × 10¹⁵
N₂(g) + 4H₂(g) +Cl₂ $$\Updownarrow$$ 2NH₄Cl(s) K_p = 3.9 × 10⁷⁰

Determine the K_p for N₂(g) + 3H₂ (g) $$\Updownarrow$$ 2NH₃(g).

Homework Equations

Don't know/none

The Attempt at a Solution

In order to get to N₂(g) + 3H₂ (g) $$\Updownarrow$$ 2NH₃(g), I would have to multiple/flip equations so that they result in the desired reaction set.
-(H₂ (g) + Cl₂(g) $$\Updownarrow$$ 2HCl (g) K_p = 2.5 × 10³³)
-2(NH₃(g) + HCl(g) $$\Updownarrow$$ NH₄Cl(s) K_p = 2.1 × 10¹⁵)
N₂(g) + 4H₂(g) +Cl₂ $$\Updownarrow$$ 2NH₄Cl(s) K_p = 3.9 × 10⁷⁰

These would cancel out to the desired reaction.

In Hess's Law, I understand that multiplying a step would mean its enthalpy gets multiplied by that number. If I flip a step, its enthalpy would inverse its sign.

In voltage calculation from standard reduction potentials, reversing the sign would inverse the potential for the step but multiplying the step does not affect the potential.

Originally I would just follow Hess's Law to calculate but I never did this for K_p and so I'm not sure the answer would be correct. How do I approach this problem and solving for K_p.

 

[Borek]

Just combine these reactions so that everything cancels out - you will be left with Kp=f(Kp1,Kp2,Kp3) (indices just to signal these are constants for each reaction given). That's all.

 

[312213]

Ok thanks.

On a side note, I think the latex reference for some of the arrows are wrong.