Chapter 12: Problem 15
Define what is meant by unimolecular and bimolecular steps. Why are termolecular steps infrequently seen in chemical reactions?
Chapter 12: Problem 15
Define what is meant by unimolecular and bimolecular steps. Why are termolecular steps infrequently seen in chemical reactions?
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Get started for freeA popular chemical demonstration is the "magic genie" procedure, in which hydrogen peroxide decomposes to water and oxygen gas with the aid of a catalyst. The activation energy of this (uncatalyzed) reaction is 70.0 \(\mathrm{kJ} / \mathrm{mol}\) . When the catalyst is added, the activation energy (at \(20 .^{\circ} \mathrm{C} )\) is 42.0 \(\mathrm{kJ} / \mathrm{mol}\) . Theoretically, to what temperature ( \((\mathrm{C})\) would one have to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at \(20 .^{\circ} \mathrm{C} ?\) Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.
The reaction $$ 0^{\circ} \mathrm{C}, $$ These relationships hold only if there is a very small amount of \(\mathrm{I}_{3}^{-}\) present. What is the rate law and the value of the rate constant? (Assume that rate $=-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{SeO}_{3}\right]}{\Delta t} )$
A first-order reaction has rate constants of $4.6 \times 10^{-2} \mathrm{s}^{-1}\( and \)8.1 \times 10^{-2} \mathrm{s}^{-1}\( at \)0^{\circ} \mathrm{C}\( and \)20 .^{\circ} \mathrm{C},$ respectively. What is the value of the activation energy?
The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.
For the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the following data were collected, where $$ \text {Rate} =-\frac{\Delta\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\Delta t} $$ Calculate \(E_{\mathrm{a}}\) for this reaction.
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