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## Homework Statement

H

_{2}(g) + Cl

_{2}(g) [tex]\Updownarrow[/tex] 2HCl (g)

*K*

_{p}= 2.5 × 10

^{33}

NH

_{3}(g) + HCl(g) [tex]\Updownarrow[/tex] NH

_{4}Cl(s)

*K*

_{p}= 2.1 × 10

^{15}

N

_{2}(g) + 4H

_{2}(g) +Cl

_{2}[tex]\Updownarrow[/tex] 2NH

_{4}Cl(s)

*K*

_{p}= 3.9 × 10

^{70}

Determine the

*K*

_{p}for N

_{2}(g) + 3H

_{2}(g) [tex]\Updownarrow[/tex] 2NH

_{3}(g).

## Homework Equations

Don't know/none

## The Attempt at a Solution

In order to get to N

_{2}(g) + 3H

_{2}(g) [tex]\Updownarrow[/tex] 2NH

_{3}(g), I would have to multiple/flip equations so that they result in the desired reaction set.

-(H

_{2}(g) + Cl

_{2}(g) [tex]\Updownarrow[/tex] 2HCl (g)

*K*

_{p}= 2.5 × 10

^{33})

-2(NH

_{3}(g) + HCl(g) [tex]\Updownarrow[/tex] NH

_{4}Cl(s)

*K*

_{p}= 2.1 × 10

^{15})

N

_{2}(g) + 4H

_{2}(g) +Cl

_{2}[tex]\Updownarrow[/tex] 2NH

_{4}Cl(s)

*K*

_{p}= 3.9 × 10

^{70}

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}.