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 4 of 4: The cold upper atmosphere

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\frac{k_B(175)}{k_B(300)} = e^{\frac{20000}{8.314}\left(\frac{1}{175}-\frac{1}{300}\right)} \approx e^{5.7} \approx 300\ \Rightarrow\ \text{Proposal B dominates}

Analysis

Proposal A's rate falls drastically at 175 K ( $e^{-80/RT}$ ), while proposal B's rate rises by roughly $e^{5.7} \approx 300$ -fold between 300 and 175 K. Since $\ce{H2SO4}$ forms faster in the upper atmosphere, proposal B — the negative- $E_a$ channel — must dominate there.