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Questions and Answers
What is a simplifying assumption that can be used when K is very small?
What is a simplifying assumption that can be used when K is very small?
- Most reactants remain unchanged at equilibrium. (correct)
- The initial concentrations must be equal.
- Products are favored and reactants are nearly depleted.
- The changes in concentrations of products are significant.
Which rule allows for simplifying assumptions when the larger value is at least 100 times greater than the smaller value?
Which rule allows for simplifying assumptions when the larger value is at least 100 times greater than the smaller value?
- The Simplification Rule
- The Square Rule
- The Doubling Rule
- The Hundred Rule (correct)
When applying the perfect square solution method, what is the primary benefit?
When applying the perfect square solution method, what is the primary benefit?
- It eliminates the need for stoichiometric coefficients.
- It speeds up the process by avoiding quadratic equations. (correct)
- It allows for a direct calculation of equilibrium concentrations.
- It avoids using any concentration values.
In the example of a reaction with K = 64, what equation is simplified when using the square root method?
In the example of a reaction with K = 64, what equation is simplified when using the square root method?
What assumption can be made about the reactants when K is significantly large?
What assumption can be made about the reactants when K is significantly large?
How can you determine if simplifying assumptions can be made in equilibrium problems?
How can you determine if simplifying assumptions can be made in equilibrium problems?
What is the resulting equilibrium concentration of [HI] when K = 64 and the initial concentrations are set up correctly?
What is the resulting equilibrium concentration of [HI] when K = 64 and the initial concentrations are set up correctly?
What general trend in the equilibrium constant (K) indicates that most of the reactants remain at equilibrium?
What general trend in the equilibrium constant (K) indicates that most of the reactants remain at equilibrium?
When can we make simplifying assumptions in chemical equilibrium problems?
When can we make simplifying assumptions in chemical equilibrium problems?
Which of the following scenarios allows for the assumption that one reactant is completely depleted in an equilibrium reaction?
Which of the following scenarios allows for the assumption that one reactant is completely depleted in an equilibrium reaction?
When is it appropriate to use an extra variable (e.g., 'y') to solve for small values in an equilibrium problem?
When is it appropriate to use an extra variable (e.g., 'y') to solve for small values in an equilibrium problem?
What should we do to solve for the equilibrium concentrations of products when a reactant is completely consumed?
What should we do to solve for the equilibrium concentrations of products when a reactant is completely consumed?
In an equilibrium reaction with a large K, which of the following is NOT a valid assumption?
In an equilibrium reaction with a large K, which of the following is NOT a valid assumption?
Flashcards
Large K value
Large K value
A significantly high equilibrium constant indicating products are favored.
Limiting reagent
Limiting reagent
The reactant that is completely consumed first in a reaction.
Simplifying assumption
Simplifying assumption
A technique used to simplify calculations in equilibrium problems.
Stoichiometric coefficient
Stoichiometric coefficient
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Equilibrium constant expression
Equilibrium constant expression
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Perfect Square Solution
Perfect Square Solution
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Large K
Large K
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Small K
Small K
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The Hundred Rule
The Hundred Rule
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The Thousand Rule
The Thousand Rule
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ICE Table
ICE Table
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Study Notes
Simplifying Assumptions in Equilibrium Constant Problems
- Simplifying assumptions speed up equilibrium constant problems, avoiding the quadratic equation.
- Perfect Square Solution: Use when the equilibrium constant expression has squared terms in both numerator and denominator.
- Simplify by taking the square root of both sides to remove squares and simplify the equation.
- Large or Small Equilibrium Constant (K): Employable when K is significantly larger or smaller than initial concentrations.
- The Hundred Rule or Thousand Rule: Use assumptions if the larger value (concentration or K) is at least 100 or 1000 times greater than the smaller value.
- K is very small: Minimal shift from initial conditions, so reactant concentrations with large starting values remain nearly unchanged.
- Ignore the change in initial concentrations of those reactants.
- K is very large: Products are highly favored.
- One or more reactants are nearly completely consumed (approximately zero) at equilibrium.
- If multiple reactants, identify limiting reagent based on stoichiometric coefficients and assume this reactant reaches zero.
Example: Perfect Square Solution
- Reaction: H2 + I2 ⇌ 2HI; K = 64
- Initial: [H2] = 0.5 mol/L, [I2] = 0.5 mol/L, [HI] = 0 mol/L
- Equilibrium expression: K = [HI]2/([H2] * [I2]) = 64 = (2x)2/(0.5 - x)2
- Simplify by taking square roots: 8 = 2x / (0.5 - x)
- Solve for x: x = 0.4 mol/L
- Equilibrium concentrations: [H2] = [I2] = 0.1 mol/L, [HI] = 0.8 mol/L
Example: Large K and Small Change
- Reaction: H2 + I2 ⇌ 2HI; K = 8.5 x 10-8
- Initial: [H2] = 0.55 mol/L, [I2] = 0.55 mol/L, [HI] = 0 mol/L
- Large ratio (0.55 / 8.5 x 10-8) suggests simplifying assumptions.
- K is small; shift to equilibrium is minimal.
- Assume negligible change in [H2] and [I2] concentrations.
- Solve for [HI] directly.
Example: Large K and Large Change
- Reaction: 2CO + O2 ⇌ 2CO2; K = 5.2 x 109
- Initial: [CO] = 0.55 mol/L, [O2] = 0.55 mol/L, [CO2] = 0 mol/L
- Large ratio (5.2 x 109 / 0.55) supports simplifying assumptions.
- K is large; products are favored.
- Assume [CO] drops to nearly zero.
- Solve for [CO2] and [O2] using the change in [CO].
Example: Large K, Large Change, with Limiting Reagent
- Reaction: 2CO + O2 ⇌ 2CO2; K = 7.5 x 108
- Initial: [CO] = 0.44 mol/L, [O2] = 0.44 mol/L, [CO2] = 0 mol/L
- Large ratio (7.5 x 108 / 0.44) allows for simplification.
- K is large and CO's stoichiometric coefficient (2) suggests CO is the limiting reagent.
- Assume [CO] drops to nearly zero.
- Solve for [CO2] and [O2] using the change in [CO].
General Strategies for Simplifying Assumptions
- ** Large/Small K:**
- Small shift (small K): Ignore changes in reactant concentrations with significant initial values.
- Large shift (large K): Reactants are essentially consumed during the shift to equilibrium; assume some reactant reaches zero.
- Limiting Reactant: If multiple reactants are involved, identify the one that is consumed completely and assume its concentration reaches zero.
- Solving for Small Values: Often involves using a separate variable (e.g., y) in the equilibrium expression to solve for a small change in concentration.
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