Which reaction is accompanied by an increase in entropy? a) zns(s) + 3/2 o2(g) → zno(s) + so2(g) b) ch4(g) + h2o(g) → co(g) + 3h2(g) c) bao(s) + co2(g) → baco3(s) d) na2co3(s) + co2(g) + h2o(g) → 2 nahco3(s) e) n2(g) + 3 h2(g) → 2 nh3(g)

B) CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase .

If in a reaction , more number of gaseous atoms are present in the product side , entropy will increase , i.e. Δ°S > 0

As the molecules in liquid state are loosely packed and has less force of attraction between the molecules, but as it is converted to solid, the force of attraction between the molecule increases and hence entropy decreases.

If in a reaction , less number of gaseous atoms are present in the product side , entropy will decrease , i.e. Δ°S < 0

From the question ,

A) ZnS(s) + 3/2 O₂(g) → ZnO(s) + SO₂(g)

Gaseous atoms -

Reactant - 0 + 3/2 = 3/2

Product - 0 + 1 = 1 ,

Less number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S < 0 , i.e. , decrease in entropy

B) CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)

Gaseous atoms -

Reactant - 1 + 1 = 2

Product - 1 + 3 = 4 ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0 , i.e. , increase in entropy .

C) BaO(s) + CO₂(g) → BaCO₃(s)

Gaseous atoms -

Reactant - 0 + 1 = 1

Product - 0  ,

Less number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S < 0 , i.e. , decrease in entropy .

D) Na₂CO₃(s) + CO₂(g) + H₂O(g) → 2 NaHCO₃(s)

Gaseous atoms -

Reactant - 0 + 1 + 1 = 2

Product = 0 ,

Less number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S < 0 , i.e. , decrease in entropy .

E) N₂(g) + 3 H₂(g) → 2 NH₃(g)

Gaseous atoms -

Reactant - 1 + 3 = 4

Product - 2 ,

Less number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S < 0 , i.e. , decrease in entropy .

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.

2b2t

explanation:

2b2t