Problem 2

Professor Molina (MIT, Nobel Prize 1995) studied the acid-rain reaction producing

\ce{H2SO4}

in the atmosphere and proposed two stoichiometries: Proposal A:

\ce{H2O(g) + SO3(g) -> H2SO4(g)}

; Proposal B:

\ce{2H2O(g) + SO3(g) -> H2SO4(g) + H2O(g)}

. (2.1) Using simple collision theory, what reaction orders are expected for A and B? Proposal B is thought to proceed via

\ce{SO3 + 2H2O <=>[k_1][k_{-1}] SO3.2H2O}

(fast) and

\ce{SO3.2H2O ->[k_2] H2SO4 + H2O}

(slow,

k_2 \ll k_1, k_{-1}

). (2.2) Apply the steady-state principle to derive the rate law and order of this mechanism. (2.3) Quantum calculations give overall activation energies

E_A = +80

kJ mol⁻¹ (A) and

E_B = -20

kJ mol⁻¹ (B). Write the Arrhenius temperature dependence for each and predict how each rate constant varies with T. (2.4)

\ce{H2SO4}

formation is faster in the upper atmosphere (175 K) than at the surface (300 K): which pathway dominates in the upper atmosphere?

Step 3 of 4: Arrhenius temperature dependence

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Intuition

A negative activation energy signals a pre-equilibrium: colder gas holds more of the reactive $\ce{SO3.2H2O}$ complex, outweighing the slower individual collisions.

k_A = A\,e^{-80/RT}\ (\text{increases with } T);\qquad k_B = A\,e^{+20/RT}\ (\text{decreases with } T)

Analysis

With $k = A e^{-E_a/RT}$ : proposal A has $E_a = +80$ kJ mol⁻¹ so $k_A$ rises with T. Proposal B's apparent $E_a = -20$ kJ mol⁻¹ (the exothermic complexation outweighs the slow-step barrier), so $k_B = A e^{+20/RT}$ rises as T falls.