8. 10/20 points | Previous Answers My Notes Consider the following equilibrium:

Question

8. 10/20 points | Previous Answers My Notes Consider the following equilibrium: 2NOBr(g) -2N0(g) + Br2(g) An equilibrium mixture is 0.104 M NOBr, 0.399 M NO, and 0.204 M Br2 a) What is the value of Kc at the temperature of the above concentrations? Kc = 3.00 b) How many moles/liter of NOBr must be added to the above equilibrium mixture to produce an equilibrium mixture that is 0.418 M Br2? 292 mol/L NOBr must be added C) If the temperature is 380 K, what is the value of Kp? Kp = 93.59 atm d) what is the value of Go at 380 K? AGo=-14339.9 Xk)

Explanation / Answer

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

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

       = 0.399^2*0.204/0.104^2

       = 3

b)

   concentration Br2 increased = 0.418-0.204 = 0.214 M

so that,

concentration of NO increased to 0.399+2*0.214 = 0.827 M

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

    3   = 0.827^2*0.418/((0.104+x)-2*0.214)^2

x = concentration of NOBr must add = 0.6327 M

c)

   Dn = np -nR

      = 3-2 = 1

Kp = Kc*(RT)^Dn

    = 3*(0.0821*380)^1

    = 93.594

d) DG = - RTlnKp

       = - 8.314*380ln93.59

       = -14339.9 joule

      = -14.34 kj

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