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: $1.1\times 10^4$

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,

$K_c=\frac{[CO_2]^3\times [H_2O]^3}{[O_2]^4\times [C_3H_6O]^1}$

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

$K_c=\frac{(1.8)^3\times (2.0)^3}{(0.30)^4\times (0.51)^1}$

$K_c=1.1\times 10^4$

Thus the value of the equilibrium constant is $1.1\times 10^4$