Question

Consider the following equilibrium: 2NOBr(g) <------> 2NO(g) + Br2(g). An equilibrium mixture is 0.146 M NOBr, 0.270 M NO, and 0.236 M Br2.

a) What is the value of Kc at the temperature of the above concentrations?

b) How many moles/liter of NOBr must be added to the above equilibrium mixture to produce an equilibrium mixture that is 0.439 M Br2?

c) If the temperature is 370 K, what is the value of Kp?

d) What is the value of Go at 370 K?

Answer 1

a)

Kc = [NO]^2[Br2] / [NOBr]^2

Kc = (0.27^2)(0.236)/(0.146^2) = 0.8071120

so

b)

in order to get Br = 0.439

find...

initially:

[NOBr] = 0.146 + x

the change:

0.439-0.236 = 0.203 for Br2

then

NO = 0.27+0.203*2= 0.676will be the equilbirium concentration

so assume "x" is the amoutn added:

Kc = [NO]^2[Br2] / [NOBr]^2

0.8071120 = (0.676^2)(0.439) / (0.146 +x)^2

solve for x

(0.146 +x)^2= 0.8071120 / ((0.676^2)(0.439) )

0.146 +x = sqrt(4.02323)

x =  sqrt(4.02323)-0.146

x = 1.85979

we must add 1.85979 M of NOBR

c)

T = 370 K

so

K p = Kc*(RT)^dn

dn = 2+1-2 = 1

Kp = Kc*(RT)

Kp = (0.8071120)(0.082*370)) = 24.487

d)

faluve of G0

G0 = -RT*ln(K)

G0 = -8.314*370*ln(0.8071120) = 659.20333 J/mol