Question

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=5.3

Gaseous hydrogen iodide is placed in a closed container at 425∘C, where it partially decomposes to hydrogen and iodine:

2HI(g)⇌H2(g)+I2(g)

At equilibrium it is found that [HI]= 3.55×10⁻³M, [H2]= 4.82×10⁻⁴M , and [I2]= 4.82×10⁻⁴M.

What is the value of Kc at this temperature?

b

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=5.3

a

Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?

b

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?

Answer 1

                A + B   ⇌ C   +   D

initially     1       2         0         0

at equi (1 - X) (2 - X) X        X

Kc=[C] [D] / [A] [B] = 5.3 = (x) *(x) / ((1-x)*(2-x))

x = 0.8726 = [D]

                 2 HI (g)      ⇌    H2 (g)   +      I2 (g)

at equi   3.55×10⁻³        4.82×10⁻⁴      4.82×10⁻⁴

Kc = [HI]^2 / [H2] [I2]

Kc = (3.55 * 10^-3)^2 / (4.82*10^-4)^2

Kc = 54.2453

   A + B   ⇌ C   +   D

initially     2       2         0         0

at equi (2- X) (2 - X) X        X

Kc=[C] [D] / [A] [B] = 5.3 = (x) *(x) / ((2-x)*(2-x))

x = 1.3943 = [D]

Answer 2

Part.a :-

The ICE table of the given reaction is :

........................A..............+............B <-----------> C...............+..................D

Initial (I)..........2.0 M......................2.0 M..............0.0 M............................0.0 M

Change (C).....- Y...........................- Y...................+ Y...............................+ Y

Equilibrium (E).(2.0 -Y) M...........(2.0 -Y).................Y M..............................Y M

Expression of equilibrium constant i.e. K_c is :

K_c = [C].[D] / [A]. [B]

5.3 = Y² /(2.0 -Y) (2.0 - Y)

5.3 = Y² / (4.0 - 4.0 Y + Y² )

5.3 (4.0 - 4.0 Y + Y² ) = Y²

Y² = 21.2 - 21.2 Y + 5.3 Y²

21.2 - 21.2 Y + 5.3 Y² - Y² = 0

4.3 Y² - 21.2 Y + 21.2= 0

On solving this quadratic equation, We have

Y = 3.54 (Which is not possible) and Y = 1.39 (Possible)

Hence,Final equilibrium concentration of A is = 2.0 - Y = 2.0 - 1.39= 0.61 M

Part.b :-

The ICE table of the given reaction is :

........................A..............+............B <-----------> C...............+..................D

Initial (I)..........1.0 M......................2.0 M..............0.0 M............................0.0 M

Change (C).....- Y...........................- Y...................+ Y...............................+ Y

Equilibrium (E).(1.0 -Y) M...........(2.0 -Y).................Y M..............................Y M

Expression of equilibrium constant i.e. K_c is :

K_c = [C].[D] / [A]. [B]

5.3 = Y² /(1.0 -Y) (2.0 - Y)

5.3 = Y² / (2.0 - 3.0 Y + Y² )

5.3 (2.0 - 3.0 Y + Y² ) = Y²

Y² = 10.6 - 15.9 Y + 5.3 Y²

4.3 Y² - 15.9 Y + 10.6 = 0

On solving this quadratic equation, We have

Y = 2.83 (Which is not possible) and Y = 0.87 (Possible)

Hence,Final equilibrium concentration of D is = Y = 0.87 M