Some reactions are second order under low reactant concentrations, but are first order athigher reactant concentrations. A mechanism used to explain this phenomenon is shownbelow. In the first step of this mechanism, a collision partner, C, collides with the reactantmolecule (R), resulting in the formation of an energize reactant molecule R*. In the secondstep, this high-energy molecule further reacts to form product, P. The collision partner maybe a second reactant molecule (another molecule of R), a product molecule (P), or an inertgas such as nitrogen or argon.Overall reaction: R → PStep 1: R (g) + C (g) ⇌ R* (g) + C (g)Step 2: R* (g) → P(a) Using the steady-state approximation for [R*], show that the rate law for the overallreaction is: . To receive full credit, you must clearly show/explain yourwork.(b) Show that the rate law is overall second order at low pressures (that is, lowconcentrations of R and C).(c) Show that the overall rate law is first order at high pressures (that is, highconcentrations of R and C).(d) Explain why the rate law for the reaction CH3 NC (g) → CH 3 CN (g) is second order inCH 3 NC (g) at low concentrations.k1 k 2 [R] [C]k−1 [C] + k 2Youmans Page of2 2k1k–1k2

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Some reactions are second order under low reactant concentrations, but are first order at
higher reactant concentrations. A mechanism used to explain this phenomenon is shown
below. In the first step of this mechanism, a collision partner, C, collides with the reactant
molecule (R), resulting in the formation of an energize reactant molecule R*. In the second
step, this high-energy molecule further reacts to form product, P. The collision partner may
be a second reactant molecule (another molecule of R), a product molecule (P), or an inert
gas such as nitrogen or argon.
Overall reaction: R → P
Step 1: R (g) + C (g) ⇌ R* (g) + C (g)
Step 2: R* (g) → P
(a) Using the steady-state approximation for [R*], show that the rate law for the overall
reaction is: . To receive full credit, you must clearly show/explain your
work.
(b) Show that the rate law is overall second order at low pressures (that is, low
concentrations of R and C).
(c) Show that the overall rate law is first order at high pressures (that is, high
concentrations of R and C).
(d) Explain why the rate law for the reaction CH3 NC (g) → CH 3 CN (g) is second order in
CH 3 NC (g) at low concentrations.
k1 k 2 [R] [C]
k−1 [C] + k 2
Youmans Page of2 2
k1
k–1
k2

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