Question

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

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:

$Kc=\frac{[CO_2]^3[H_2O]^3}{[C_3H_6O][O_2]^4}$

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

$Kc=\frac{[1.8]^3[2.0]^3}{[0.51][0.30]^4}$

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