For the reaction- 2 SO 2 g - O 2 g - 2 SO 3 g - the entropy-

The correct option is C decreasesFor the reaction, #160; 2 SO _ 2 ( g ) + O _ 2 ( g ) → 2 SO _ 3 ( g ) ,In this case, there are 2 moles of a gas on the p ...

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Byju's Answer

Standard XII

Chemistry

Entropy

For the react...

Question

For the reaction,
$2 {S O}_{2} (g) + O_{2} (g) \to 2 {S O}_{3} (g)$ , the entropy:

increases

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decreases

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remains unchanged

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change cannot be predicted

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Solution

The correct option is C decreases
For the reaction,
$2 {S O}_{2} (g) + O_{2} (g) \to 2 {S O}_{3} (g)$ ,
In this case, there are 2 moles of a gas on the product side and 3 moles of gas on the reactant side. The reaction results in a decrease in the amount of gas, so the entropy change is negative.

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Similar questions

Q. Predict the conditions for forward reaction on the basis of Le Chatelier's principle for :

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

;

Δ H = - 198 k J

2 S O_{2} (g) + O_{2} (g) ⟶ 2 S O_{3} (g)

For this reaction relate

Δ H

and

Δ U

Q. The conditions favourable for the reaction:

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g); △ H^{\circ} = - 198 K J

are

Q. The conditions favourable for the reaction:

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g); △ H^{o} = - 198 k J

are :

Q. Consider the reaction :

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g); △ H^{o} = - 198 k J

The favourable conditions for the forward reaction are:

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For the reaction,2SO2(g)+O2(g)→2SO3(g), the entropy:

The correct option is C decreasesFor the reaction, 2SO2(g)+O2(g)→2SO3(g),In this case, there are 2 moles of a gas on the product side and 3 moles of gas on the reactant side. The reaction results in a decrease in the amount of gas, so the entropy change is negative.

For the reaction,
$2 {S O}_{2} (g) + O_{2} (g) \to 2 {S O}_{3} (g)$ , the entropy:

The correct option is C decreases
For the reaction,
$2 {S O}_{2} (g) + O_{2} (g) \to 2 {S O}_{3} (g)$ ,
In this case, there are 2 moles of a gas on the product side and 3 moles of gas on the reactant side. The reaction results in a decrease in the amount of gas, so the entropy change is negative.