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 1 of 4: Orders from collision theory

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\text{A:}\ \frac{d[\ce{H2SO4}]}{dt} = k[\ce{H2O}][\ce{SO3}]\ (2^{nd}\ \text{order});\qquad \text{B:}\ \frac{d[\ce{H2SO4}]}{dt} = k[\ce{SO3}][\ce{H2O}]^2\ (3^{rd}\ \text{order})

Analysis

In simple collision theory the rate is proportional to the collision frequency: a bimolecular step gives order 2, a termolecular step (two $\ce{H2O}$ + one $\ce{SO3}$ ) gives order 3.