equilibrium homework help

Question

If Kp = 7.3 × 10⁻⁶ for the reaction below at 400.0 K, then what is the value of Kc? Note Kc is sometimes called K. (R = 0.0821 L･atm/mol･K.) 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g)

Answer 1

Excellent answer provided by @Alex M., but if you actually want to know how that was obtained, see below.

The equation we can use is Kp = Kc(RT)^∆n

∆n = change in number of moles = 4 moles product - 2 mols reactant = 2

7.3x10^-6 = Kc(0.0821 x 400)² = 1078 Kc

Kc = 7.3x10^-6 / 1078

Kc = 6.8x10^-9 (to 2 sig. figs.)

Answer 2

6.78 x 10^-9 (no units on Kc values).