Q7(d). Second-order rate law with unequal concentrations; units of k
(i) Consider a second-order reaction A + B → P, where the initial concentration of A is ‘a’ mol dm⁻³ and that of B is ‘b’ mol dm⁻³. After time ‘t’, x mol dm⁻³ of A and x mol dm⁻³ of B react to form x mol dm⁻³ of the product P. Show that the second-order rate constant for this reaction will be given by k₂ = [1/((a − b)t)] ln[b(a − x)/a(b − x)] with the assumption that a > b. What will be the unit of k₂? (Consider time in seconds.) (ii) Determine the units of the rate constants for zeroth-order and 5/2-order reactions. Assume that concentrations are expressed in mol dm⁻³ and time in seconds.
Answer in brief
(i) k₂ = [1/((a − b)t)] ln[b(a − x)/a(b − x)], unit dm³ mol⁻¹ s⁻¹. (ii) Zero order: mol dm⁻³ s⁻¹; order 5/2: dm9/2 mol−3/2 s⁻¹.
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Coverssecond-order kineticsintegrated rate lawpartial fractionsunits of rate constantorder of reactionisolation method
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