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 2 of 4: Steady-state rate law

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\frac{d[\ce{SO3.2H2O}]}{dt} = k_1[\ce{SO3}][\ce{H2O}]^2 - (k_{-1} + k_2)[\ce{SO3.2H2O}] = 0\ \Rightarrow\ v = k_2[\ce{SO3.2H2O}] = \frac{k_1 k_2}{k_{-1} + k_2}[\ce{SO3}][\ce{H2O}]^2 \approx K_{eq}k_2[\ce{SO3}][\ce{H2O}]^2

Analysis

Setting the complex's net rate of formation to zero gives $[\ce{SO3.2H2O}] = k_1[\ce{SO3}][\ce{H2O}]^2/(k_{-1}+k_2)$ . Since $k_2 \ll k_{-1}$ , the pre-equilibrium is essentially undisturbed and the overall rate is third order with effective constant $K_{eq}k_2$ .