Question

Consider the following reaction and associated equilibrium constant: aA(g)⇌bB(g), Kc = 4.0

Part A Find the equilibrium concentrations of A and B for a=1 and b=1. Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction. Express your answers using two significant figures separated by a comma. [A], [B] =   M   Part B Find the equilibrium concentrations of A and B for a=2 and b=2. Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction. Express your answers using two significant figures separated by a comma. [A], [B] =   M   Part C Find the equilibrium concentrations of A and B for a = 2 and b = 1. Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction. Express your answers using two significant figures separated by a comma. [A], [B] =   M

Answer 1

aA(g)bB(g)

For above reaction, gas equilibrium constant is given by

K_c=([C]^c[D]^d)/([A]^a[B]^b)

Part A

a=1, b=1

A(g)B(g)

Initial concentration 1.0M 0

Change -x +x

Equilibrium concentration 1.0-x x

So Kc=[B]/[A]

4.0=x/(1.0-x)

4.0(1.0-x)=x

4.0-4x=x

4.0=5x

x=4/5=0.8M

So [A]=1.0-x=1.0-0.8=0.20M

[B]=x=0.80 M

Part B

a=2, b=2

2A(g)2B(g)

Initial concentration 1.0M 0

Change -2x +2x

Equilibrium concentration 1.0-2x 2x

So Kc=[B]²/[A]²

4.0=(2x)²/(1.0-2x)²

4.0(1.0-2x)²=4x²

4.0(1+ 4x²-4x)= 4x²

4.0+16x²-16x=4x²

4+16x²-16x-4x²= 0

12x²-16x+4=0

12x²-12x-4x+4=0

12x(x-1)-4(x-1)=0

(12x-4)(x-1)=0

So x=4/12 or x=1

Now x cannot be 1.0M so x=4/12=1/3 M=0.33M

So [A]=1.0-x=1.0-0.33=0.67M

[B]=x=0.33 M

Part C

a=2, b=1

2A(g)B(g)

Initial concentration 1.0M 0

Change -2x +x

Equilibrium concentration 1.0-2x x

So Kc=[B]²/[A]²

4.0=x²/(1.0-2x)²

4.0(1.0-2x)²=x²

4.0(1+ 4x²-4x)= x²

4.0+16x²-16x=x²

4+16x²-16x-x²= 0

15x²-16x+4=0

15x²-10x-6x+4=0

5x(3x-2)-2(3x-2)=0

(5x-2)(3x-2)=0

So x=2/5 or 2/3

x=0.4 M or 0.66M

x cannot be 0.66M because then [A] will be 1.0-2x0.66M=1.0-1.22M=-0.22M which is not possible

So [A]=1.0-2x=1.0-2x0.4=1-0.8=0.2M

[B]=x=0.4M