Chemistry, 08.03.2020 23:37 ELGuapo6746

Describe a reversible reaction using the sample equation below. Include forward reaction and reverse reaction in your discussion.
2SO2 (g) + O2(g) <--> 2SO3 (g)

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Answer from: shira35

The equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the products and the product of the equilibrium concentrations of the reactants, all raised to the power of their respective stoichiometric coefficients in the chemical equation.

For the given equation,

2SO2(g] + O2(g] ⇌ 2SO3(g]

This reaction has two reactants, SO2 and O2, and one product, SO3. This tells that the equilibrium constant will feature three terms.

The equilibrium constant for the forward reaction will be,

Kc = $\frac{[SO3]^{2}}{[SO2]^2. [O2]}$

We can write equilibrium constant for the reverse reaction K′c, using the same approach,

2SO2(g] + O2(g] ⇌ 2SO3(g]

The reaction has one product and two reactants, so the equilibrium constant for the reverse reaction will be,

K'c = $\frac{[SO2]^2. [O2]}{[SO3]^{2}}$

Remember,

K'c = 1 / Kc

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

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

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