Question

For the following equilibrium:

2A+B⇌2C

If initial concentrations are [A]=0.80 M,[B]=0.95 M,[C]=2.5 M, and at equilibrium [C]=1.9 M, what is the equilibrium constant?

need help please show me how to solve it, thanks

Answer 1

Answer:

1.5 M

Explanation:

We can determine the equilibrium constant for

2A+B⇌2C

by using the initial and equilibrium concentrations of C to determine the unknown equilibrium concentrations of A and B.

First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −2x as the change in concentration of C. (Note that we know the change in concentration will be negative because the equilibrium concentration for C is smaller than the initial concentration. The coefficient is 2 because two moles of C are produced for every two moles of A and one mole of B that react.)

                                                     A         B    C  

 Initial Concentration                 0.80      0.95   2.5  

 Change                                           2x        x        −2x  

 Equilibrium Concentration  0.80+2x   0.95+x   1.9  

We can now calculate the value of x and the equilibrium concentrations of the products.

                                           A   B      C  

 Initial Concentration       0.80 0.95    2.5  

 Change                               0.60 0.30   −0.60  

 Equilibrium Concentration  1.40 1.25       1.9  

Now we can calculate the equilibrium constant.

Kc=1.9^2/(1.40^2)(1.25)

Kc=1.473

The answer should have two significant figures, so round to 1.5.

Answer 2

Given: Initial concentration of A = 0.80M

           Initial concentration of B = 0.95M

           Initial concentration of C = 2.5M

           At equilibrium C = 1.9M

To find: The equilibrium constant.

Solution:

Two moles of A reacts with one mole of B to form two moles of C.

                                      A               B                 C

Initial concentration     0.80         0.95           2.5

Change                         2x                x           2x    

At equlibrium               1.9              0.95          1.9

Now, Kc = (1.9)²/(1.9)²(0.95)

              = 1.05    

Therefore, the equlibrium constant will be 1.05.

• In a chemical reaction, chemical equilibrium is the state in which both the reactants and products are present in concentrations that have no further tendency to change with time so that there is no observable change in the properties of the system.