The equilibrium constant in terms of pressures, K_p, for the reaction NH_3(g) +

Question

The equilibrium constant in terms of pressures, K_p, for the reaction NH_3(g) + H_2Sec(g) NH_4HSe(s) at 25 degree C is 1.18 times 10^4. (a) If the partial pressure of ammonia is P_NH_3 = 0.617 atm and solid ammonium hydrogen selenide is present, what is the equilibrium partial pressure of hydrogen selenide at 25 degree C? P_H_2 Se = atm (b) An excess of solid NH_4 at 25 degree C and a pressure of 2.46 times 10^-2 atm. Calculate the pressures of NH_3(g) and H_2Se(g) reached at equilibrium. P_NH_3 = atm P_H_2Se = atm

Explanation / Answer

a)

NH3 (g) + H2Se <—> NH4HSe (s)

0.617       0                   (initial)

0.617+x       x                   (at equilibrium)

Kp = 1/(p(NH3)*p(H2Se))

1.18*10^4 = 1/((0.617+x)*x)

1.18*10^4 = 1/(0.617*x + x^2)

7.281*10^3*x + 1.18*10^4 *x^2 = 1

11800*x^2 + 7281*x -1 = 0

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

a = 11800

b = 7281

c = -1

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 = 53060161

roots are :

x = 2.75*10^-4 and x = -0.617

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

x = 1.37*10^-4

x is nothing but pressure of H2Se

Answer: 1.37*10^-4

b)

NH3 (g)        +   H2Se <—> NH4HSe (s)

2.46*10^-2           0                   (initial)

2.46*10^-2+x           x                   (at equilibrium)

Kp = 1/(p(NH3)*p(H2Se))

1.18*10^4 = 1/((2.46*10^-2+x)*x)

1.18*10^4 = 1/(2.46*10^-2 *x+x^2)

290.28*x + 1.18*10^4 * x^2 = 1

1.18*10^4 * x^2 + 290.28*x - 1 = 0

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

a = 11800

b = 290.28

c = -1

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 = 131462.5

roots are :

x = 3.06*10^-3 and x = -0.028

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

x = 3.06*10^-3

p(NH3) = 2.46*10^-2 - x = 2.46*10^-2 - 3.06*10^-3 = 2.15*10^-2 atm

p(H2Se) = x = 3.06*10^-3 atm

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