Reaction Mechanisms and Catalysis: Quiz

Reaction Mechanisms and Catalysis: Quiz

7 questions

1. Question 1EasyWhat is the rate law of the elementary step 2NO₂ → N₂O₄?

   Arate = k[NO₂]Brate = k[NO₂]²Crate = k[N₂O₄]DIt cannot be written without experiments
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   Answer: rate = k[NO₂]²

   For an elementary step the rate law follows the coefficients: two NO₂ molecules collide, so the step is second order in NO₂.

2. Question 2EasyIn the mechanism Cl + O₃ → ClO + O₂, then ClO + O → Cl + O₂, what is Cl?

   AAn intermediateBA catalystCA productDA transition state
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   Answer: A catalyst

   Cl is used in step 1 and regenerated in step 2. ClO, made then used up, is the intermediate.

3. Question 3MediumFor 2NO₂Cl → 2NO₂ + Cl₂, step 1 (slow) is NO₂Cl → NO₂ + Cl. What is the rate law?

   Arate = k[NO₂Cl]²Brate = k[NO₂Cl]Crate = k[NO₂Cl][Cl]Drate = k[NO₂]²[Cl₂]
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   Answer: rate = k[NO₂Cl]

   The slow step is unimolecular, so the reaction is first order in NO₂Cl, even though the overall equation has a coefficient of 2.

4. Question 4HardA fast equilibrium NO + Br₂ ⇌ NOBr₂ is followed by the slow step NOBr₂ + NO → 2NOBr. What is the rate law?

   Arate = k[NOBr₂][NO]Brate = k[NO][Br₂]Crate = k[NO]²[Br₂]Drate = k[NO]²
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   Answer: rate = k[NO]²[Br₂]

   Rate = k₂[NOBr₂][NO], and [NOBr₂] = K₁[NO][Br₂] from the pre-equilibrium, so rate = k₂K₁[NO]²[Br₂].

5. Question 5MediumAn enzyme has Vmax = 80 µmol L⁻¹ min⁻¹ and Km = 4.0 mmol/L. What is the rate at [S] = 4.0 mmol/L?

   A20 µmol L⁻¹ min⁻¹B40 µmol L⁻¹ min⁻¹C60 µmol L⁻¹ min⁻¹D80 µmol L⁻¹ min⁻¹
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   Answer: 40 µmol L⁻¹ min⁻¹

   When [S] = Km, v = ½Vmax = ½ × 80 µmol L⁻¹ min⁻¹ = 40 µmol L⁻¹ min⁻¹.

6. Question 6MediumAdding much more substrate restores the full rate of an inhibited enzyme. What type of inhibitor is present?

   ACompetitiveBNon-competitiveCA catalyst poison that binds permanentlyDNone: the enzyme was denatured
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   Answer: Competitive

   A competitive inhibitor competes for the active site, so a large excess of substrate outcompetes it and the rate approaches the same Vmax.

7. Question 7EasyWhy does an enzyme-catalysed reaction slow down above its optimum temperature?

   AThe activation energy risesBMolecules collide less oftenCThe enzyme denatures and loses the shape of its active siteDThe equilibrium constant falls
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   Answer: The enzyme denatures and loses the shape of its active site

   Heat disrupts the weak interactions that hold the protein in shape, so the active site no longer fits the substrate.