[Boxxxed]

1.  The decomposition of NH4I is an endothermic process.  At 25oC, experiment shows that the total pressure of the gases in equilibrium with the solid is 1 atm:

NH4I (s) → NH3 (g) + HI (g)

a. Find the equilibrium pressures of NH3 and HI. (3 points)
b. Find Kp for this reaction at 25oC. (2 points)


For a) would the pressures be 0.5 atm for NH3 and HI?

How would I start b). Just sub in Kp=Kc(RT)^dn?

Or would Kp be (0.5)(0.5)=0.25atm?

[DevaDevil]

For a) would the pressures be 0.5 atm for NH3 and HI?

correct, if the reactant was 100% NH₄I (in vacuum), which decomposed into NH₃ and HI, then you have formed both products in equal amounts, so each will have a partial pressure of 0.5 atm (assuming the solid has no contribution in vapor form)

Or would Kp be (0.5)(0.5)=0.25atm?

yes it would be (just check the units of K_p, which will be atm²), find again the definition of reaction equilibrium constants in partial pressures.

[Boxxxed]

Next part of the problem

e. Calculate the equilibrium pressures of NH3(g) and HI(g) if 2 atm of NH3 is suddenly added to the reaction flask. (11 points)


Kp=0.25=(2.50-x)(0.50-x)

x=0.382

Would this be the correct way to solve this?

[opti384]

Yes but keep in mind that your final answer will not be 0.382 

[Boxxxed]

Yes I have to subtract it from both pressure values

[DevaDevil]

indeed, and do not forget to answer with the appropriate units!