For the reaction below, Kc = 0.60 at 550 K. The reaction starts with a 0.050 M c

Question

For the reaction below, Kc = 0.60 at 550 K. The reaction starts with a 0.050 M concentration of A_2B_2, at 550 K. A_2B_2 (g) 2AB (g) You must use the quadratic formula to determine the equilibrium concentrations of each species; the small x approximation will not work. You will report specific information from your solution to the problem in 33-36. Equilibrium constant (Kc) expression for the reaction A) Kc = [2 AB]^2/[A_2 B_2] B) Kc = [AB]^2/[A_2B_2] C) Kc = [2AB]/[A_2B_2] D) Kc = [A_2B_2]/[AB]^2 E) Kc = [A_2B_2]/[2AB] Expression for the product from equilibrium row of ICE table A) x B) 2x C) -x D) -2x E) x^2 Magnitude of "c" from quadratic formula (i.e., sign does not matter) A) 4 B) 0.60 C) 0.03 D) 0.10 E) 2 [A_2B_2]_eq A) 0.0396 M B) 0.0302 M C) 0.0104 M D) 0.0213 M

Explanation / Answer

33.A2B2(g)<------------>2AB(g)

Equilibrium Constant:

Kc = [AB]^2/[A2B2]

34.A2B2<----------->2AB

Initially,0. 050M. 0

At equilibrium,

(0.05-x). 2x

Option (B) is correct.

35.Kc= [AB]^2/[A2B2]

= (2x)^2/(0.05-x)

0.60 =4x^2/(0.05-x)

0.6(0.05-x)=4x^2

4x^2+0.6x-0.03=0

Compare with ax2+bx+c=0

c= -0.03

Option (c) is correct.

36.4x^2+0.6x-0.03=0

Compare with ax2+bx+c=0

By Sridharacharya formula get the value of x.

And find [A2B2]eq =(0.05-x)

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