Consider the following equilibrium: An equilibrium mixture is 0.167 M NOBr, 0.20

Question

Consider the following equilibrium:

An equilibrium mixture is 0.167 M NOBr, 0.209 M NO, and 0.179 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 0.353 M Br2?

c) If the temperature is 380 K, what is the value of Kp?

Consider the following equilibrium: 2NOBr(g) 2NO(g) + Br2(g) An equilibrium mixture is 0.167 M NOBr, 0.209 M NO, and 0.179 M Br2. What is the value of Kc at the temperature of the above concentrations? Kc = M How many moles/liter of NOBr must be added to the above equilibrium mixture to produce an equilibrium mixture that is 0.353 M Br2? If the temperature is 380 K, what is the value of Kp? Kp = What is the value of delta G degree at 380 K?

Explanation / Answer

(a) Kc = [NO]^2*[Br2]/ [NOBr]^2

= [0.209]^2[0.179]/[0.167]^2

= 0.280 M

(b) 2NOBr(g) ---> 2NO (g) + Br2 (g)

strat 0.167+X 0.209 0.179

equi 0.167+X-2Y 0.209+2Y 0.179

0.353 => Y = 0.353-0.179 = 0.174

X-0.181 0.557 0.353

Kc = 0.28 = (0.557)^2*0.353 / (X-0.181)^2

X = 0.806 M

(c) Kp = Kc * (RT)^ (2+1-2) = 0.28* 0.082*380 = 8.724 atm

(d dG^o =- RT*lnKp = -8.314 *380*ln(8.724) = -6849.8 J

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