1.Consider the following unbalanced reaction: HCl (g) ? H2 (g) + Cl2 (g) a.If th

Question

1.Consider the following unbalanced reaction: HCl (g) ? H2 (g) + Cl2 (g) a.If the equilibrium constant of this reaction is Kp = 4.17*10-34, determine the equilibrium constant for H2 (g) + Cl2 (g) ? HCl (g) at 25 oC. b.If this were an endothermic reaction, how would increasing the temperature alter the equilibrium constant? c.Given 1.00 atm Cl2, 0.567 atm H2 and 2.50 atm HCl, determine their equilibrium partial pressures at 25 oC 1.Consider the following unbalanced reaction: HCl (g) ? H2 (g) + Cl2 (g) a.If the equilibrium constant of this reaction is Kp = 4.17*10-34, determine the equilibrium constant for H2 (g) + Cl2 (g) ? HCl (g) at 25 oC. b.If this were an endothermic reaction, how would increasing the temperature alter the equilibrium constant? c.Given 1.00 atm Cl2, 0.567 atm H2 and 2.50 atm HCl, determine their equilibrium partial pressures at 25 oC 1.Consider the following unbalanced reaction: HCl (g) ? H2 (g) + Cl2 (g) a.If the equilibrium constant of this reaction is Kp = 4.17*10-34, determine the equilibrium constant for H2 (g) + Cl2 (g) ? HCl (g) at 25 oC. b.If this were an endothermic reaction, how would increasing the temperature alter the equilibrium constant? c.Given 1.00 atm Cl2, 0.567 atm H2 and 2.50 atm HCl, determine their equilibrium partial pressures at 25 oC

Explanation / Answer

1.

(a)

HCl (g) H2 (g) + Cl2 (g)

Kp = 4.17 x 10-34

Now,

H2 (g) + Cl2 (g) HCl (g)

The equilibrium expression written for a reaction written in the reverse direction is the reciprocal of the one for the forward reaction.

Kp' = 1/Kp
= 1 / (4.17 x 10-34)
= 2.40 x 1033

(b)

heat + HCl (g) H2 (g) + Cl2 (g) (endothermic reaction)

Raising the temperature favors the forward reaction (endothermic).
Lowering the temperature favors the reverse reaction (exothermic).

When the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant.

(c)

  HCl (g) H2 (g) + Cl2 (g)
IP: 2.5 0.567 1.00
C: - x + x + x
EP: 2.5 - x 0.567 + x 1.00 + x

Kp = [ (0.567 + x) (1 + x) ] / (2.5 - x)
or, 4.17 x 10-34 = [ (0.567 + x) (1 + x) ] / (2.5 - x)
or, (4.17 x 10-34) (2.5 - x) = (0.567 + 0.567x + x + x2)  
or, 1.04 x 10-33 - (4.17 x 10-34)x = (0.567 + 0.567x + x + x2)  
or, 1.04 x 10-33 - (4.17 x 10-34)x = 0.567 + 1.567x + x2
or, x2 + 1.567x + (4.17 x 10-34)x + 0.567 - (1.04 x 10-33)  = 0
or, x2 +  1.567x + 0.567 = 0

solving the quadratic equation, we get
x = -0.566 and x = -1

discarding the value of x = -1, since x value can't be greater than initial pressure, so we have x = - 0.566.

Hence, the partial pressures are
HCl = 2.5 - x = 2.5 - (-0.566) = 3.066 atm
H2 = 0.567 + x = 0.567 - 0.566 = 0.001 atm
Cl2 = 1.00 + x = 1 - 0.566 = 0.434 atm

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