PLEASE HELP DUE ON MONDAY, KINLY SHOW ALL WORKINGS ND CALCULATIONS!!! 7. Compoun

Question

PLEASE HELP DUE ON MONDAY, KINLY SHOW ALL WORKINGS ND CALCULATIONS!!!

7. Compounds A, B, and C react according to the following equation:

                        3A(g)   +   2B(g)   <=>    2C(g)

At 100 oC a mixture of these gases at equilibrium showed that [A] = 0.855 M, [B] = 1.23 M, and [C] = 1.75
What is the value of Kc for this reaction?

8. A mixture 0.500 mole of carbon monoxide and 0.400 mole of bromine was placed into a rigid 1.00-L container and the system was allowed to come to equilibrium. The equilibrium concentration of COBr2 was 0.233 M. What is the value of Kc for this reaction?

CO(g)   +   Br2(g)   <=>   COBr2(g).

Explanation / Answer

7.

The Kc for the given reaction is:

Kc = [C]^2 / [A]^3 [B]^2

Take note that:

* the products are on the numerator

* the reactants are on the denominator

* the products and reactants are raised to their respective stoichiometric (i.e. accdg to the chemical equation) coefficients

* the concentrations indicated are the concentrations at equilibrium

So it is just a matter of substituting the given values:

Kc = (1.75)^2 / (0.855)^3 (1.23)^2

Kc = 3.24

8. at equilibrium
[CO]= 0.500 - 0.233 =0.267 M
[Br2]= 0.400- 0.233=0.167 M
Kc = 0.233/ 0.267 x 0.167=5.23

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