## Quiz 3 Preparation #8 atm^2 for Kp

RaquelAvalos1K
Posts: 35
Joined: Fri Sep 26, 2014 2:02 pm

### Quiz 3 Preparation #8 atm^2 for Kp

#8 Has the value of Kp at .11atm2
Since Kp is unitless, and there is a unit here, and there are no concentrations/partial pressures for the reactants or products, can this problem not be solved?

Thank you!

Eduardo Torres 3G
Posts: 13
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Quiz 3 Preparation #8 atm^2 for Kp

Although the equilibrium constants are unit less, they are a result of multiplying values with units. The problem is trying to let you know that your final answer will in units of atm and it can be solved. Since you start off with pure NH4HS(s), and NH3(g) and H2S(g) have a mole ration of 1 to 1, the products will be produced by the reaction at the same rate. This makes the value of Kp=(Partial Pressure of NH3)(Partial Pressure of H2S). Since the products are being produced at the same rate their partial pressures will be equal so this can be rewritten as Kp=x^2. You take the square root of both sides and there is you answer.

RaquelAvalos1K
Posts: 35
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Quiz 3 Preparation #8 atm^2 for Kp

Okay, and atm being written with a power of 2 is the same as atm? And since atm2 is saying the answer is in atm, there is no conversion regarding bar, yea?

About the 1:1 rate ratio, you are referring to the change in composition for the I.C.E table, correct? As in what the stoichiometric coefficients are? And we are only considering the ratio of the products since they are gases and the reactant is a solid, so irrelevant?

If it was 1NH3(g) + 2H2S(g), the Change would read: +1X +2X across, and we would have Kp=2x2, yea?

Thank you so much!

Eduardo Torres 3G
Posts: 13
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Quiz 3 Preparation #8 atm^2 for Kp

Yes I believe you got it.