9. The reaction (CH3)3CBr + OH-1 ? (CH3)3COH + Br-1 in a certain solvent is firs

Question

9. The reaction
(CH3)3CBr + OH-1 ? (CH3)3COH + Br-1

in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH-1. In several experiments, the rate constant k was determined at different temperatures. A plot of ln (k) versus 1/T was constructed resulting in a straight line with a slope of -1.10 x 104 K and y-intercept of 33.5. Assume k has units of s-1.

A. Determine the activation energy for this reaction.
B. Determine the value of the frequency factor A.
C. Calculate the value of k at 25

Explanation / Answer

Firstly, we need to consider the rate order with respect to reactants of the reaction: (CH3)3CBr(aq) + OH-(aq) ? (CH3)3COH(aq) + Br-(aq) In observing the experimental results: ____|_[(CH3 )3CBr]_|_[OH-]_|_Initial Rate_| Exp. |___(mol/L)____|_(mol/L)|__(mol/L.s) _| 1._____0.10_________0.10___1.0 x 10-3 2._____0.20_________0.10___2.0 x 10-3 3._____0.10_________0.20___1.0 x 10-3 4._____0.30_________0.20_____? In comparison expt 1 to expt 2, doubling the [(CH3 )3CBr] (while keeping [OH-] constant) doubles the reaction rate. =>Thus, the reaction is first order with respect to [(CH3 )3CBr] In comparison expt 1 to expt 3, doubling [OH-] (while keeping [(CH3 )3CBr] constant) doesn't change the reaction rate. => Thus, the reaction is zeroth order with respect to [OH-] Overall reaction rate order = 1 rate law can be expressed as Rate = k [(CH3 )3CBr]], where k is rate constant. As rate constant for a reaction is unchanged at constant temperature, Then, for expt 1, 1.0E-3 = k *(0.10) ----(1) for expt 2, 2.0E-3 = k* (0.20) ----(2) Subtracting (2) with (1), => 1.0E-3 = k*(0.10) => k = 1.0E-2 s^-1

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