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DENIUS [597]
4 years ago
5

Be sure to answer all parts. Consider the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g)ΔH = −198.2 kJ/mol How would the concentrations of S

O2, O2, and SO3 at equilibrium change if we were to (a) increase the temperature? [SO2]: The concentration increases. The concentration decreases. There is no change. [O2]: The concentration increases. The concentration decreases. There is no change. [SO3]: The concentration increases. The concentration decreases. There is no change. (b) increase the pressure and decrease the volume? [SO2]: The concentration increases. The concentration decreases. There is no change. [O2]: The concentration increases. The concentration decreases. There is no change. [SO3]: The concentration increases. The concentration decreases. There is no change. (c) add a catalyst? [SO2]: The concentration increases. The concentration decreases. There is no change. [O2]: The concentration increases. The concentration decreases. There is no change. [SO3]: The concentration increases. The concentration decreases. There is no change.
Chemistry
1 answer:
andrew-mc [135]4 years ago
4 0

Answer:

a)

[SO2]: The concentration increases

[O2]: The concentration increases.

[SO3]: The concentration decreases.

b)

[SO2]: The concentration decreases.

[O2]: The concentration decreases.

[SO3]: The concentration increases.

c)

[SO2]:There is no change.

[O2]: There is no change.

[SO3]: There is no change

Explanation:

For an exothermic reaction, increase in temperature decreases the concentration of products and increases the concentration of reactants since increase in temperature shifts the equilibrium position to the left hand side.

Increase in pressure and decrease in volume will shift the equilibrium position towards the right hand side which means more SO3 in the system.

Catalyst increases the rate of forward and reverse reaction simultaneously hence at equilibrium, the concentration of reactants and products remain unchanged.

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Answer:

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Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen
Oksanka [162]
<h3>Answer:</h3>

10.6 mol NO

<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
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O

[Given] 13.2 mol O₂

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

[RxN] 5 mol O₂ → 4 mol NO

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 13.2 \ mol \ O_2(\frac{4 \ mol \ NO}{5 \ mol \ O_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 10.56 \ mol \ NO

<u>Step 4: Check</u>

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

10.56 mol NO ≈ 10.6 mol NO

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