A sample of carbonyl bromide is sealed in a 150.0-cm^3 glass bulb to which a pre

Question

A sample of carbonyl bromide is sealed in a 150.0-cm^3 glass bulb to which a pressure gauge is attached The bulb is heated to 73.0 degree C, and the gauge shows that the pressure in the bulb rises to 0.883 atm. At this temperature the COBr_2(g) is partially dissociated into co(g) and Br_2(g) according to the equation COBr_2(g) Equilibrium CO(g) + Br_2(g) At 73.0degree C. K_p = 5.40 for this reaction. Assume that the contents of the bulb are at equilibrium and calculate the partial pressure of the three different chemical species in the vessel. P_COBr_2 = _____ atm P_CO = _____ atm P_Br_2 = _____ atm

Explanation / Answer

Let the initial pressure of only CoBr2 be p atm

COBr2 (g)   <—> CO (g) + Br2 (g)

p               0           0       (initial)

p-x               x           x       (at equilibrium)

Given, total pressure = 0.883

so,

p-x +x+x = 0.883

p = 0.883-x

Kp = p(CO)*p(Br2)/p(COBr2)

5.40 = x*x / (p-x)

5.40 = x*x / (0.883-2x)

4.768 - 10.8*x = x^2

x^2 + 10.8*x - 4.768 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1.0

b = 10.8

c = -4.768

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 135.712

roots are :

x = 0.425 and x = -11.225

since x can't be negative, the possible value of x is

x = 0.425

SO,

p(COBr2) = 0.883-2x = 0.883 - 2*0.425 = 0.033 atm

p(CO) = x = 0.425 atm

p(Br2) = x = 0.425 atm

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