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2 years ago

Consider the following reversible reaction.

Chemistry

1 answer:

katovenus [111]2 years ago

4 0

The equilibrium constant for the reaction is:

K subscript eq equals StartFraction StartBracket upper H subscript 2 EndBracket superscript 2 StartBracket upper O subscript 2 EndBracket over StartBracket upper H subscript 2 upper O EndBracket superscript 2 EndFraction.

<h3>What is equilibrium constant? </h3>

The equilibrium constant (Kₑq) for a given reaction is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

For example, the equilibrium constant Kₑq for the reaction below is given as

2A <=> B

Kₑq = [B]/[A]²

<h3>How to determine the equilibrium constant </h3>

2H₂O(g) <=> 2H₂(g) + O₂(g)
Equilibrium constant (Kₑq) =?

Kₑq = [H₂]²[O₂] / [H₂O]²

K subscript eq equals StartFraction StartBracket upper H subscript 2 EndBracket superscript 2 StartBracket upper O subscript 2 EndBracket over StartBracket upper H subscript 2 upper O EndBracket superscript 2 EndFraction.

Learn more about equilibrium constant:

brainly.com/question/17960050

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<h3>Answer:</h3>

5.55 mol C₂H₅OH

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Tables
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

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

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

<u>Step 4: Check</u>

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

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

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