Experiment shows that the rate of formation of carbon tetrachloride from chlorof

Question

Experiment shows that the rate of formation of carbon tetrachloride from chloroform, CHCI_3(g) + Cl_2(g) rightarrow CCl_4(g) + HCI(g), is first order in CHCI, and 1/2, order in cl_2. Determine which is the slow step from the following elementary steps. Show your work (i.e. solve for the rate law using the slow step). All steps are shown with equilibrium back and forth arrows, though one is the slow step. Cl_2(g) 2CI(g) CI(g)+CHCl_3(g) HCI(g)+CCI_3(g) CCI_3(g)+CI(g) CCl_4 (g) Write the experimental rate law (i.e. in the form of rate = k[A]"[B]"') for this reaction. Write the rate law for each elementary step. Identify the intermediates. Write the backwards rate law if necessary. Using the information in parts a and b, identify the rate determining (slow) step for this reaction. Justify your choice by solving for the rate law of the slow step with no intermediates.

Explanation / Answer

a) Experimental rate law,

rate = k [CHCl3]1 [Cl2]1/2

b) step 1: rate = k[Cl2]

step 2: rate = k [Cl] [CHCl3]

step 3 : rate = k [CCl3] [Cl]

c) step 2 is the rate determining (slow) step of the reaction.

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