A and B react as follows: A +
A and B react as follows: A + B = C + D. The equilibriumconstant is 2.0 x 10^3. If 0.30 mol of A and 0.80 mol of B aremixed in 1 L, what are the concentrations for A, B, C, and D afterreaction?
Answer:
Since volume is 1 L, number of moles is same as molarity
Let’s prepare the ICE table
[A] [B] [C] [D]
initial 0.3 0.8 0 0
change -1x -1x +1x +1x
equilibrium 0.3-1x 0.8-1x +1x +1x
Equilibrium constant expression is
Kc = [C]*[D]/[A]*[B]
2000.0 = (1*x)(1*x)/((0.3-1*x)(0.8-1*x))
2000.0 = (1*x^2)/(0.24-1.1*x + 1*x^2)
480-2200*x + 2000*x^2 = 1*x^2
480-2200*x + 1999*x^2 = 0
Let’s solve this quadratic equation
Comparing it with general form: (ax^2+bx+c=0)
a = 1.999*10^3
b = -2.2*10^3
c = 4.8*10^2
solution of quadratic equation is found by below formula
x = {-b + √(b^2-4*a*c)}/2a
x = {-b – √(b^2-4*a*c)}/2a
b^2-4*a*c = 1.002*10^6
putting value of d, solution can be written as:
x = {2.2*10^3 + √(1.002*10^6)}/3.998*10^3
x = {2.2*10^3 – √(1.002*10^6)}/3.998*10^3
solutions are :
x = 0.8006 and x = 0.2999
x can’t be 0.8006 as this will make the concentrationnegative.so,
x = 0.2999
At equilibrium:
[A] = 0.3-1x = 0.3-1*0.29991 = 0.00009 M
[B] = 0.8-1x = 0.8-1*0.29991 = 0.50009 M
[C] = 0+1x = 0+1*0.29991 = 0.29991 M
[D] = 0+1x = 0+1*0.29991 = 0.29991 M
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