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denpristay [2]
3 years ago
14

Which situation would cause the following equilibrium reaction to decrease the formation of the products? 2SO2 (g) + O2 (g) Two

arrows stacked on top of each other. The top arrow points to the right. The bottom arrow points to the left. 2SO3 (g) + Energy
increase the temperature
decrease the volume
increase the pressure
decrease the temperature
Chemistry
2 answers:
vovangra [49]3 years ago
8 0

The given equilibrium reaction is,

2 SO_{2} (g) + O_{2}(g)  2 SO_{3} (g) + Energy

The given reaction is exothermic. So, heat energy will be a product. Therefore, decreasing the temperature (heat energy) would lead to the formation of more products as when the amount of energy which is a product is reduced, there is more room for the products to form.

Increasing the pressure would shift the equilibrium towards that side which has least number of moles of the gaseous substance. Hence, here increasing the pressure would lead to the formation of more products by shifting the equilibrium towards the right side.

Decreasing the volume would make the equilibrium shift towards the least number of moles of the gaseous substance. So, here in this equilibrium decreasing the volume would lead to the formation of more products.

MaRussiya [10]3 years ago
8 0

Answer:  increase the temperature

Explanation: Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

2SO_2+O_2\leftrightharpoons 2SO_3+energy

This is a type of Exothermic reaction because heat is released in the reaction.

Option A: Increase the temperature

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in temperature occurs. As, this is an exothermic reaction, forward reaction will increase the temperature. Hence, the equilibrium will shift in the left direction, products will decrease.

Option B: Decrease the volume  and Option C: increase in pressure

If the volume of the container is decreased, the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is greater at the reactant side. So, the equilibrium will shift in the right direction, products will increase.

Option D: Decrease the temperature

If the temperature is decreased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs. As, this is an exothermic reaction, forward reaction will increase the temperature. Hence, the equilibrium will shift in the right direction, products will increase.

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Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

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Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

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