Question

What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.51 M

C3H6O, 0.30 M O2, 1.8 M CO2, and 2.0 M H2O?

C3H6O(g)+4O2(g)\rightleftharpoon3CO2(g)+3H2O(g)

Answer 1

Answer:- Kc = 11294

Solution:- The given balanced equation is:

`C_3H_6O(g)+4O_2(g)\rightleftharpoons 3CO_2(g)+3H_2O(g)`

Let's write the equilibrium expression for this:

`Kc=([CO_2]^3[H_2O]^3)/([C_3H_6O][O_2]^4)`

Let's plug in the values of given equilibrium concentrations and do the calculations:

`Kc=([1.8]^3[2.0]^3)/([0.51][0.30]^4)`

Kc = 11294

So, the numerical value of Kc is 11294.

Answer 2

The numerical value of Kc is 1.129 x10^4

Explanation

C3H6O +4O2→ 3 CO2 + 3H2O

KC is the ratio of concentration of the product over the reactant.

Each concentration of product and reactant are raised to the power of its coefficient.

Therefore the KC expression of equation above is

Kc=[ (Co2)^3 (H2O)^3] / [(C3H6O) (O2)^4)]

Kc =[(1.8^3) x (2.0^3)] / [(0.51) x (0.30^4)] =1.129 x10^4