Question

Part A 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)⇌3CO2(g)+3H2O(g)

A) 2.4 × 101B) 1.1 × 104C) 8.9 × 10-5D) 4.3 × 10-2

Answer 1

Answer:
`1.1* 10^4`

Step-by-step explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
`K_c`.

The given balanced equilibrium reaction is,

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

At eqm. conc. (0.51) M (0.30) M (1.8) M (2.0)M

The expression for equilibrium constant for this reaction will be,

`K_c=([CO_2]^3* [H_2O]^3)/([O_2]^4* [C_3H_6O]^1)`

Now put all the given values in this expression, we get :

`K_c=((1.8)^3* (2.0)^3)/((0.30)^4* (0.51)^1)`

`K_c=1.1* 10^4`

Thus the value of the equilibrium constant is
`1.1* 10^4`