Consider the following reaction: A(g)<-->2B(g) Find the equilibrium partial pres

Question

Consider the following reaction: A(g)<-->2B(g)
Find the equilibrium partial pressures of A and B for each ofthe following different values of Kp. Assume that theinitial partial pressure of B in each case is 1.0 atm and that theinitial partial pressure of A is 0.0 atm. Make any appropriatesimplifying assumptions. Find answers for both equilibriums (A andB) a)Kp = 1.4 b)Kp= 2.0 x 10-4 c)Kp= 2.0 x 105 Consider the following reaction: A(g)<-->2B(g)
Find the equilibrium partial pressures of A and B for each ofthe following different values of Kp. Assume that theinitial partial pressure of B in each case is 1.0 atm and that theinitial partial pressure of A is 0.0 atm. Make any appropriatesimplifying assumptions. Find answers for both equilibriums (A andB) a)Kp = 1.4 b)Kp= 2.0 x 10-4 c)Kp= 2.0 x 105

Explanation / Answer

                                             2B(g) <----> A (g) initialpressure                         1atm             0 change                                   -x                 + x/2 Equbpressure                       1-x                  x/2 Euilibriumconstant K = p A / p 2B (1) Kp = 1.4 = (x/2) / (1-x)^2             1.4( 1-x)^2 = x/2            2.8(1-x)^2 = x             2.8(1+x^2- 2x) = x           2.8 +2.8x^2 - 11.2x = x           2.8 + 2.8x^2 - 11.2x - x = 0          2.8x^2 -12.2x +2.8 = 0    x= [(-12.2) ±( (-12.2)^2 -4(2.8)(2.8))] / (*2.8)      On solving for x we get x = 0.2428atm & x = 4.1142 atm x/2 = 4.1142 / 2 = 2.0572 atm which is not taken since it is> 1 atm ( pressure of B) So we take x = 0.2428 So the Equb pressure of A = x/2 = 0.1214 atm Eub. pressure of B = 1-x = 0.7572 atm Simillarly do the next two (1) Kp = 1.4 = (x/2) / (1-x)^2             1.4( 1-x)^2 = x/2            2.8(1-x)^2 = x             2.8(1+x^2- 2x) = x           2.8 +2.8x^2 - 11.2x = x           2.8 + 2.8x^2 - 11.2x - x = 0          2.8x^2 -12.2x +2.8 = 0    x= [(-12.2) ±( (-12.2)^2 -4(2.8)(2.8))] / (*2.8)      On solving for x we get x = 0.2428atm & x = 4.1142 atm x/2 = 4.1142 / 2 = 2.0572 atm which is not taken since it is> 1 atm ( pressure of B) So we take x = 0.2428 So the Equb pressure of A = x/2 = 0.1214 atm Eub. pressure of B = 1-x = 0.7572 atm Simillarly do the next two

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