Determine the rate law for the elementary reaction 2I(g) + M(g) rightarrow^k I_2

Question

Determine the rate law for the elementary reaction 2I(g) + M(g) rightarrow^k I_2(g) + M(g), where M is any molecule present in the reaction container besides atomic iodine and iodine. What are the units of the rate constant k for this reaction? Is this reaction identical to 2I(g) rightarrow^k I^2(g). Explain. Consider the reaction mechanism A + 13 C - with the forward rate constant k_1 and the reverse rate constant k_1' - and C rightarrow P with the rate constant k_2. Write the expression for d[P] / dt under the following assumptions: steady state approximation pre-equilibrium Show that the mechanism paraH_2(p) 2H(g) (Step 1) and H(g) + paraH_2(g) rightarrow orthoH_2(g) + H(g) (Step 2) is consistent with the rate law d[orthoH_2] / dt = k_obs [paraH_2]^3/2 if Step 1 is a fast equilibrium (i.e.. under pre-equilibrium conditions). What is k_obs in terms of the rate constants for each step? Note: Use the same nomenclature we have been using in class to express rate constants for each step.

Explanation / Answer

The rate law for the elementary reaction 2I (g) +M -k- > I2 (g)+ M (g) is following:

rate = k[A]^2

Here M is present in the reaction beside atomic I and I2 but not participate in the reaction.

This reaction is example of second order reaction.The unit of rate cosnat is k is folloing:

L mole-1 sec-1

d[I] in mol / L / dt in sec = k[mol / L]^2

k= L mole-1 sec-1

Hence the this reaction is identicle with the following reaction:

2I (g) + -k- > I2 (g)

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