The following reactions have the indicated equilibrium constants at a particular temperature: n2(g) + o2(g) ⇌ 2no(g)kc = 4.3 × 10−25 2no(g) + o2(g) ⇌ 2no2(g)kc = 6.4 × 109 determine the value of the equilibrium constant for the following equation at the same temperature: n2(g) + 2o2(g) ⇌ 2no2(g) kc = × 10 (enter your answer in scientific notation.)

Explanation:

The given equilibrium reaction is:

The expression for the equilibrium constant is:

The another given equilibrium reaction is:

The expression for the equilibrium constant is:

To find,

For the equilibrium which is:

The expression for the equilibrium constant is:

Multiplying and dividing by and rearranging in the above equation as:

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.