A catalyst increases reaction rate by

Answer: Lowering Eₐ via an alternate mechanism Catalysts provide a lower-Eₐ pathway; ΔH and equilibrium are unchanged.

What does the catalyzed line's shallower slope indicate?

Answer: A lower activation energy. Slope = −Eₐ/R. Shallower (less negative) slope means smaller Eₐ — the catalyst lowers the activation energy.

What is the effect of a catalyst on the reaction?

Answer: Lowers Ea, increases rate A catalyst provides an alternate pathway with a lower Ea, increasing rate without changing ΔH.

AP Chemistry · Topic 5.11

Catalysis Practice

Part of Kinetics.(TRA-5.E)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$A catalyst increases reaction rate by$
- A
  $Raising T$
- B
  $Increasing ΔH$
- C
  $Shifting equilibrium toward products$
- D
  $Lowering Eₐ via an alternate mechanism$
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Why
$Catalysts provide a lower-Eₐ pathway; ΔH and equilibrium are unchanged.$
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Sample 2difficulty 2/5
$The same reaction is run with and without a catalyst, with k measured at varied T.$
$What does the catalyzed line's shallower slope indicate?$
- A
  $A lower activation energy.$
  check_circle
- B
  $A first-order reaction.$
- C
  $A larger frequency factor only.$
- D
  $A higher activation energy.$
Why
$Slope = -Eₐ/R. Shallower (less negative) slope means smaller Eₐ — the catalyst lowers the activation energy.$
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Sample 3difficulty 2/5
$Two energy profiles for the same reaction are shown: with and without a catalyst.$
$What is the effect of a catalyst on the reaction?$
- A
  $Raises Ea, slows rate$
- B
  $Changes ΔH, shifts equilibrium$
- C
  $Lowers Ea but does not change rate$
- D
  $Lowers Ea, increases rate$
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Why
$A catalyst provides an alternate pathway with a lower Ea, increasing rate without changing ΔH.$
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Sample 4difficulty 2/5
$Compared to the uncatalyzed pathway, the catalyzed pathway has:$
- A
  $Higher Ea but lower ΔH$
- B
  $Same Ea and shifted ΔH$
- C
  $Lower Ea and shifted equilibrium$
- D
  $Lower Ea but same ΔH and same K_eq$
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Why
$A catalyst speeds both forward and reverse reactions equally; ΔH and K_eq are unchanged.$
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Sample 5difficulty 2/5
$A heterogeneous catalyst$
- A
  $Cannot be regenerated$
- B
  $Is in the same phase as reactants$
- C
  $Is consumed in the reaction$
- D
  $Is in a different phase from reactants$
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Why
$Heterogeneous = different phase (e.g., solid catalyst, gas reactants).$
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