Question

Consider the reaction: 2no(g)+br2(g)⇌2nobr(g) kp=28.4 at 298 k in a reaction mixture at equilibrium, the partial pressure of no is 102 torr and that of br2 is 160 torr . part a what is the partial pressure of nobr in this mixture? express your answer using three significant figures.

Answer 1

From the Equilibrium constant we have K_p = (P_Nobr)^2/(P_No)^2 * (P_Br) 
K_p = 28.4. P_Br = 160. P_Nobr = x P_No = 102 
So we have 28.4 = x^2/ 160 * 102 
x = âš 28.4 * 160 * (102)^2 = âš47275776 = 6875.738 
To 3 Sf becomes 6880.000

Answer 2

Answer:

The partial pressure of NOBr is 43.0 torr

Explanation:

The reaction is :

2NOg)+Br₂(g)⇌2NOBr(g)

Kp = 28.4

the expression for Kp is

$Kp=\frac{[pNOBr]^{2} }{[pNO]^{2}[pBr_{2}]}$

Putting values

$28.4=\frac{[pNOBr]^{2}}{[102]^{2}[160]}$

pNOBr = partial pressure of NOBr = 42.973 torr

As mentioned we have to round of the answer to three significant figures.

So the answer is 43.0 torr.