For the overall reaction #2X_2Y_2 + Z_2 -> 2X_2Y_2Z#, which of the following mechanisms is consistent with the rate law #r(t) = k[X_2Y_2][Z_2]#?

#A)#

#Z_2 -> 2Z# (slow)
#2Z + 2X_2Y_2 -> 2X_2Y_2Z# (fast)

#B)#

#X_2Y_2 + Z_2 -> X_2Y_2Z + Z# (slow)
#X_2Y_2 + Z -> X_2Y_2Z# (fast)

#C)#

#X_2Y_2 + Z -> X_2Y_2Z# (slow)
#X_2Y_2Z + Z -> X_2Y_2Z_2# (fast)

#D)#

#X_4Y_4 + Z_2 -> X_4Y_4Z + Z# (slow)
#X_2Y_2 + X_4Y_4Z -> X_2Y_2Z + X_4Y_4# (fast)

#E)#

#X_2Y_2 + Z_2 -> X_2Y_2Z + Z# (slow)
#X_2Y_2Z -> X_2Y_2 + Z# (fast)

2 Answers
SCooke
Oct 6, 2017

A: It is the only one consistent with a first-order dependence on BOTH reactants.

Explanation:

B, C and D don't show how the second reactant changes (#Z_2 -> Z#). E is incomplete but shows the same likely problem.

Truong-Son N.
Oct 6, 2017

It's #B#. It's the only one that has a bimolecular slow step (indicating first-order for each reactant) while also cancelling out to give the overall reaction.

In a mechanism, the rate law can be approximated as the reactants in the slow step raised to their coefficients, since the slow step dominates the reaction time.

The Hess's Law-style cancellation can be seen:

#X_2Y_2 + Z_2 -> X_2Y_2Z + cancelZ# (slow)
#X_2Y_2 + cancelZ -> X_2Y_2Z# (fast)
#"----------------------------------------------"#
#2X_2Y_2 + Z_2 -> 2X_2Y_2Z#

  • #A# is incorrect because the slow step suggests a rate law of #r(t) = k[Z_2]#.
  • #C# is incorrect because it doesn't cancel out to give the overall reaction.
  • #D# is incorrect because the slow step doesn't contain #X_2Y_2#, and suggests a rate law of #r(t) = k[X_4Y_4][Z_2]#.
  • #E# is incorrect because it doesn't contain the product, and the overall reaction is just a simple dissociation.
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