The reversible chemical reaction A+B -> C+D has the following equilibrium consta

Question

The reversible chemical reaction

A+B -> C+D
has the following equilibrium constant:

Kc= [C][D]/[A][B] =3.1

Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?
Express the molar concentration concentration numerically using two significant figures.

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

Express the molar concentration numerically using two significant.

Explanation / Answer

The equilibirum constant can be defined as

Kc= [C][D]/[A][B] =3.1

A + B --> C +D

Intial conc of A , B , C and D are 2 M, @M , 0 , 0 respectively

let x amount of conc of A and B get dissociated to give x conc of C and D at equilibirum

so

Kc = [x] [x] / [a-x][a-x] = 3.1

Where a = 2M

On solving the above expression

x = 1.25

Thus final conc of A = 2-1.24 = 0.76 M

If intial conc of [A] = 1 and [B] = 2

then

Kc = [x] [x] / [1-x][2-x] = 3.1

x =0.82

therfore [D] = 0.82M

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