Question

Consider the following equilibrium: 2NOBr(g) 2NO(g) + Br2(g) An equilibrium mixture is 0.150 M NOBr, 0.395 M NO, and 0.221 M Br2. a) What is the value of Kc at the temperature of the above concentrations? Kc = M b) How many moles/liter of NOBr must be added to the above equilibrium mixture to produce an equilibrium mixture that is 0.448 M Br2? mol/L NOBr must be added c) If the temperature is 400 K, what is the value of Kp? Kp = atm d) What is the value of Go at 400 K? Go = kJ

Answer 1

2NOBr(g) 2NO(g) + Br₂(g)

a. K_c = [NO]²[Br₂]/[NOBr]²

Given:

[NO] = 0.395 M

[Br₂] = 0.221 M

[NOBr] =  0.150 M

Kc = (0.395)²(0.221)/(0.150)²

= 1.5325

Given:

Kc = 1.5325

b. [NO] = 0.395 M

[Br₂] = 0.448 M

[NOBr] =  ?

substitute these values on above equation we get

K_c = [NO]²[Br₂]/[NOBr]²

1.5325 = (0.395)²(0.448)/[NOBr] ²

[NOBr]² = 0.0456

[NOBr] = 0.2135

We know that Molarity (M) = moles/L

Hence 0.2135 moles/L of NOBr must be added to the above equilibrium mixture to produce an equilibrium mixture that is 0.448 M Br2

c. We know that Kp = Kc(RT)^Δn

  Here

R = Gas constant

     = 8.314 J/K

     = 0.0821 atm.L/mol.K

T = Temperature

     = 400K

Δn = moles of products - moles of reactant

      = 3-2

      = 1

Kc = 1.5325

subtitute these values on above equation

Kp = 1.5325*(0.0821*400)¹

        = 50.32 atm

d. We know that:

ΔG = -RTlnKp

Here:

     R = Gas constant

         = 8.314 J/K

T = 400 K

K_p =  50.32 atm

∴ ΔG = -(8.314 J/K)(400K)ln(50.32 atm)

            = 13031 J

            = 13.031 KJ