Chemistry, 22.11.2019 13:31 Playboycxm

Need with this problem on equilibrium constants

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The answer is 1.3 x 10^-29.
Essentially you are using Hess's Law with K values. In order to get the final equation, you must multiply the coefficients of the first equation by 2. Therefore, what you do to the K value is you raise the constant to 2. So you have, 5.3^2.
Next, you know you need to flip the second equation in order to obtain the final equation, but you aren't multiplying the coefficients, you don't need to. So you raise the K value to the -1 power. So you do, (2.1 x 10^30)^-1.
Multiply the two K values together to get your final equilibrium constant for the final equation.
(5.3^2) x (2.1 x 10^30)^-1
Hope that helped!

Answer from: Quest

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( $n_2$ ) .

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.

Answer from: Quest

The answer options is d

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