what is the value of k for the reaction

Question

10.00 mol of gaseous NOBr is placed into a one liter flask at 27 c, the equilibrium concentration of NO is 2.00 mol/L. what is the value of k for the reaction 2NOBr(g) <=> 2NO(g) + Br2(g) 

Answer 1

Probably the easiest way to do this is set up an ICE table, use concentrations given, solve for Kc and convert to Kp assuming that's what you are looking for.  The question says to solve for K (not k), and one assumes that is Kp, but it could be Kc.  Either way, proceed as follows:

 

2NOBr(g)   <=>   2NO(g)  +  Br₂(g)

10.00...................0..............0..........I

-2x.....................+2x...........+x........C

Since given the concentration of NO at equilibrium is 2.00, x = 1 

This makes [NOBr] at equilibrium 10.00 - 2.00 = 8

[Br₂] at equilibrium = 1

[NO] at equilibrium = 2

 

K_c = [NO]²[Br_2]/[NOBr]² = (2)²(1)/(8)² = 0.0625

K_p = K_c(RT)^Δn  = (0.0625)[(0.0821)(300)]¹  =  1.54