## Table 11.2

Hyein Cha 2I
Posts: 103
Joined: Fri Sep 29, 2017 7:05 am
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### Table 11.2

Why are thevalues of K and Kc different for some of the reactions in table 11.2? Shouldnt they all be the same?

Minie 1G
Posts: 63
Joined: Fri Sep 29, 2017 7:04 am

### Re: Table 11.2

K is Kp, the equilibrium constant based on the partial pressures. The relationship between Kc and Kp is Kp = Kc(RT)^delta(n). This can be derived from the ideal gas equation (PV=nRT) by letting P = nRT/V (and n/V is the concentration, so this is P = [gas]RT) and plugging that in to the Kp equation (the products/reactants one) and simplifying.

Anyways, we see that since R and T are the same (R is a constant and the K given is for a particular temperature) it just depends on delta(n), which is the change in moles between the products and reactants. For the first couple rows on the table, this is just 0 since there are equal moles of products and reactants. And anything raised to the 0 power is just 1, and anything multiplied by 1 doesn't change and thus Kc = Kp. But further down the table, like F2(g) <=> 2F(g) the difference in moles is 1 (2 moles F minus 1 mol F2 = 1)... so that makes Kp and Kc different by a factor of RT.

The same logic follows for the other different values :) just count the difference in moles to know if Kc and Kp are same or different.

Shreya Ramineni 2L
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Table 11.2

K=Kp which is for gases and partial pressures but Kc is for molarity.

Jessica Nunez 1I
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am

### Re: Table 11.2

To add on, K = Kp is for gases and it is denoted by the units atm (bar, torr, Pa etc.). Kc is for concentration which is denoted by molarity units.

Suhail Zaveri
Posts: 24
Joined: Sat Jul 22, 2017 3:01 am

### Re: Table 11.2

Hello,

There are many K’s even ones for dissolving called Ksp which is for dissolving. Also It is different than Kc also Kp is different as there is a calculation to convert Kc to Kp so they can be different.

Hope this helps