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

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

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

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]

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

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]

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

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