Amixture initially contains a, b, and c in the following concentrations: [a] = 0.700 m , [b] = 0.850 m , and [c] = 0.300 m . the following reaction occurs and equilibrium is established: a+2b⇌c at equilibrium, [a] = 0.570 m and [c] = 0.430 m . calculate the value of the equilibrium constant, kc.

The value of Equilibrium constant is  2.16

Explanation:

Given-

From the chemical reaction ,

ICE table can be written as -

                              A         +        2B       ⇄        C

initial moles       0.700           0.850              0.300

At equilibrium    0.700 - x     0.850 - 2x      0.300 + x

From question , at equilibrium the concentration of A = 0.570 M

The concentration of A ( ICE table ) = concentration (given in question )

0.700 - x  = 0.570

x = 0.700 - 0.570

x = 0.13

Putting the value of x in ICE table , to obtain the concentration of B

[B] = 0.850 - 2x

[B] = 0.850 - 2*0.13

[B] = 0.850 - 0.26

[B] = 0.59 M

[C] = 0.430 M       ( Given )

Equilibrium Constant ( Kc )

The value of equilibrium constant is given as the concentration of the  products each raised to the power of their respective stoichiometry by the concentration of reactants each raised to the power of their respective stoichiometry.

Kc = [C] / [A] [B]²

putting the value of concentration terms at equilibrium ,

Kc = [0.430] / [0.570][0.59]²

Kc = 2.16

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.