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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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Citrus2011 [14]
<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

<u>Step 2: Identify Conversions</u>

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

8 0
3 years ago
For the following reaction, 4.53 grams of calcium hydroxide are mixed with excess sulfuric acid. The reaction yields 6.75 grams
vagabundo [1.1K]

Answer:

\large \boxed{\text{8.32 g; 81.1 \%}}}

Explanation:

MM:                      74.09         136.14

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m/g:                      4.53  

1. Theoretical yield

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\text{Moles of Ca(OH)${_2}$} = \text{4.53 g Ca(OH)${_2}$} \times \dfrac{\text{1 mol Ca(OH)${_2}$}}{\text{74.09 g Ca(OH)${_2}$}} = \text{0.161 14 mol Ca(OH)${_2}$}

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2. Percent yield

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\text{The percent yield is } \large \boxed{\textbf{81.1 \%}}}

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Answer:

The answer to your question is:

Explanation:

Which formulas are incorrect:

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and since

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rewriting and rearranging the prior equation to make the variable V the subject, yields

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