Question

For the reaction given below, 2.00 moles of A and 3.00 moles of B are placed in a 6.00-L container. A(g) + 2B(g) C(g) At equilibrium, the concentration of A is 0.232 mol/L. What is the value of K?
a. 0.232
b. 1.15
c. 0.437
d. 1.47
e. 4.94

Answer 1

Answer:

The value for K is 4.94 (option e)

Explanation:

               A(g) +  2B (g)   →  C(g)

Initially     2m       3m              -

We have 2 moles of A and 3 moles of B

Some amount has reacted (x)

React         x         2x              x

Be careful, because ratio between A and B is 1:2

So we have the double of what we react

What happens at the equilibrium?

We have [A] in M = 0.232 mol/L

( 2moles (initially) - x)  / 6L  =  0.232 mol/L

2 - x = 0.232 m .6

2 - x = 1.392

x = 2 - 1.392 → 0.608

(x) The amount of what reacted.

                  A(g)     +      2B (g)   →      C(g)

Initially       2m                3m                  -

React       0.608           1.216             0.608

What happens at the equilibrium

                   A(g)     +      2B (g)   →         C(g)

Initially       2m                3m                     -

React       0.608             1.216               0.608

Eq             0.232        ( 3 - 1.216) /6       0.608/6

                0.232           0.297                0.101

[B] and [C] are /6, because they are not molarity.

If we make react 0.608 moles, B has to react with the double of those moles, by stoichiometry. In equilibrium we have, moles at initially - moles reacted.

Ratio between A and C is 1:1. Between B and C is 2:1

Let's make the expression for K

K = [C] / [A] . [B]²

K = 0.101 / 0.232 . 0.298²

K = 4.9

(If you use all the decimal, you will get 4.94) :), enjoy!