Steady-State and Pre-Equilibrium Approximations · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Removing the middleman from the rate law

Science · Chemistry · ages 20-21
Name ______________________   Date ____________

  1. The steady-state approximation sets which quantity to zero?

    • The concentration of the reactants
    • The intermediate's net change rate
    • The rate constant of the slow step

  2. Which species is an intermediate?

    • A species made in one step and used up in a later step
    • A catalyst added at the start and recovered at the end
    • The slowest reactant in the balanced equation

  3. The overall reaction can only go as fast as its slowest step allows.

    Circle one:   True   False

  4. Step 1 is A to I with rate k1 times [A]. Step 2 is I to P with rate k2 times [I]. Under steady state, what is [I]?

    • k1 times [A] divided by k2
    • k2 times [I] divided by k1
    • k1 times k2 times [A]

  5. When is the pre-equilibrium approximation the fair choice?

    • When the intermediate is consumed so fast that nothing balances
    • When the first step balances quickly before the intermediate is used
    • When the intermediate concentration can be measured directly

  6. The first step A to I is a fast equilibrium and the second step is slow. How do you remove [I]?

    • Set the total concentration of A to zero
    • Drop the second step from the mechanism
    • Replace [I] with an equilibrium expression in [A]

  7. A student derives a rate law from a mechanism and stops without checking data. What should they do?

    • Compare the predicted law with measured data
    • Rewrite the balanced equation until its orders match
    • Assume the mechanism is correct because the algebra closed

  8. Both approximations give the same rate law for a mechanism. What must be true of the constants?

    • The backward rate of step 1 dwarfs the forward rate of step 2
    • The forward rate of step 2 dwarfs every other constant
    • All rate constants must be exactly equal
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Removing the middleman from the rate law W1-mt_9ESWVl28-3-s1

  1. The intermediate's net change rate · Formation balances consumption, so the net change vanishes.
  2. A species made in one step and used up in a later step · Intermediates appear mid mechanism and vanish before the products.
  3. True · The slowest step is the bottleneck that sets the pace.
  4. k1 times [A] divided by k2 · Formation minus consumption is zero, so k1[A] equals k2[I].
  5. When the first step balances quickly before the intermediate is used · A fast first balance lets you swap the intermediate for an equilibrium expression.
  6. Replace [I] with an equilibrium expression in [A] · Fast balance means the equilibrium relation holds for the first step.
  7. Compare the predicted law with measured data · A mechanism earns trust only when its law matches measurement.
  8. The backward rate of step 1 dwarfs the forward rate of step 2 · That imbalance is exactly what lets the first step balance.
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