QuestionHow can you increase product formation of $\mathrm{SO}_{2}$ in the reaction: $4558 \mathrm{~kJ}+2 \mathrm{SO}_{3} \rightleftharpoons 2 \mathrm{SO}_{2}+\mathrm{O}_{2}$ ? Options: A. cool B. add $\mathrm{O}_{2}$ C. remove $\mathrm{O}$ D. remove $\mathrm{SO}_{3}$

Studdy Solution

STEP 1

Assumptions1. The given reaction is an equilibrium reaction, which means the rate of the forward reaction equals the rate of the reverse reaction. . The reaction is exothermic, as indicated by the negative sign in front of the energy term (4558 kJ).
3. Le Chatelier's principle applies, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

STEP 2

We need to determine which method could shift the equilibrium to the right, which would result in more $\mathrm{}_{2}$ being produced. According to Le Chatelier's principle, we can shift the equilibrium by changing the temperature, pressure, volume, or concentration of the reactants or products.

STEP 3

Option A suggests cooling the system. In an exothermic reaction, cooling the system will shift the equilibrium to the right because it will counteract the decrease in temperature by producing more heat. This will result in more $\mathrm{}_{2}$ being produced.

STEP 4

Option B suggests adding $\mathrm{}_{2}$ . According to Le Chatelier's principle, adding more of a product will shift the equilibrium to the left to counteract the increase in product. This will result in less $\mathrm{}_{2}$ being produced.

STEP 5

Option C suggests removing $\mathrm{}$ . However, there is no $\mathrm{}$ in the reaction, so this option is not applicable.

STEP 6

Option D suggests removing $\mathrm{}_{3}$ . According to Le Chatelier's principle, removing a reactant will shift the equilibrium to the left to counteract the decrease in reactant. This will result in less $\mathrm{}_{2}$ being produced.

STEP 7

Based on the analysis of the options, the best method to encourage more product, $\mathrm{}_{2}$ , to form from the reaction is to cool the system (Option A).

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QuestionHow can you increase product formation of $\mathrm{SO}_{2}$ in the reaction: $4558 \mathrm{~kJ}+2 \mathrm{SO}_{3} \rightleftharpoons 2 \mathrm{SO}_{2}+\mathrm{O}_{2}$ ? Options: A. cool B. add $\mathrm{O}_{2}$ C. remove $\mathrm{O}$ D. remove $\mathrm{SO}_{3}$

Studdy Solution

STEP 1

STEP 2

We need to determine which method could shift the equilibrium to the right, which would result in more $\mathrm{}_{2}$ being produced. According to Le Chatelier's principle, we can shift the equilibrium by changing the temperature, pressure, volume, or concentration of the reactants or products.

STEP 3

Option A suggests cooling the system. In an exothermic reaction, cooling the system will shift the equilibrium to the right because it will counteract the decrease in temperature by producing more heat. This will result in more $\mathrm{}_{2}$ being produced.

STEP 4

Option B suggests adding $\mathrm{}_{2}$ . According to Le Chatelier's principle, adding more of a product will shift the equilibrium to the left to counteract the increase in product. This will result in less $\mathrm{}_{2}$ being produced.

STEP 5

Option C suggests removing $\mathrm{}$ . However, there is no $\mathrm{}$ in the reaction, so this option is not applicable.

STEP 6

Option D suggests removing $\mathrm{}_{3}$ . According to Le Chatelier's principle, removing a reactant will shift the equilibrium to the left to counteract the decrease in reactant. This will result in less $\mathrm{}_{2}$ being produced.

STEP 7

Based on the analysis of the options, the best method to encourage more product, $\mathrm{}_{2}$ , to form from the reaction is to cool the system (Option A).

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Studdy Solution

STEP 1

STEP 2

We need to determine which method could shift the equilibrium to the right, which would result in more 2\mathrm{}_{2}2​ being produced. According to Le Chatelier's principle, we can shift the equilibrium by changing the temperature, pressure, volume, or concentration of the reactants or products.

STEP 3

Option A suggests cooling the system. In an exothermic reaction, cooling the system will shift the equilibrium to the right because it will counteract the decrease in temperature by producing more heat. This will result in more 2\mathrm{}_{2}2​ being produced.

STEP 4

Option B suggests adding 2\mathrm{}_{2}2​. According to Le Chatelier's principle, adding more of a product will shift the equilibrium to the left to counteract the increase in product. This will result in less 2\mathrm{}_{2}2​ being produced.

STEP 5

Option C suggests removing \mathrm{}. However, there is no \mathrm{} in the reaction, so this option is not applicable.

STEP 6

Option D suggests removing 3\mathrm{}_{3}3​. According to Le Chatelier's principle, removing a reactant will shift the equilibrium to the left to counteract the decrease in reactant. This will result in less 2\mathrm{}_{2}2​ being produced.

STEP 7

Based on the analysis of the options, the best method to encourage more product, 2\mathrm{}_{2}2​, to form from the reaction is to cool the system (Option A).

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We need to determine which method could shift the equilibrium to the right, which would result in more $\mathrm{}_{2}$ being produced. According to Le Chatelier's principle, we can shift the equilibrium by changing the temperature, pressure, volume, or concentration of the reactants or products.

Option A suggests cooling the system. In an exothermic reaction, cooling the system will shift the equilibrium to the right because it will counteract the decrease in temperature by producing more heat. This will result in more $\mathrm{}_{2}$ being produced.

Option B suggests adding $\mathrm{}_{2}$ . According to Le Chatelier's principle, adding more of a product will shift the equilibrium to the left to counteract the increase in product. This will result in less $\mathrm{}_{2}$ being produced.

Option C suggests removing $\mathrm{}$ . However, there is no $\mathrm{}$ in the reaction, so this option is not applicable.

Option D suggests removing $\mathrm{}_{3}$ . According to Le Chatelier's principle, removing a reactant will shift the equilibrium to the left to counteract the decrease in reactant. This will result in less $\mathrm{}_{2}$ being produced.

Based on the analysis of the options, the best method to encourage more product, $\mathrm{}_{2}$ , to form from the reaction is to cool the system (Option A).