Chapter 11: Problem 21
Enzymes are kinetically important for many of the complex reactions necessary for plant and animal life to exist. However, only a tiny amount of any particular enzyme is required for these complex reactions to occur. Explain.
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Chapter 11: Problem 21
Enzymes are kinetically important for many of the complex reactions necessary for plant and animal life to exist. However, only a tiny amount of any particular enzyme is required for these complex reactions to occur. Explain.
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What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law
Consider the reaction $$3 \mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E}$$ where the rate law is defined as $$-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}]$$ An experiment is carried out where \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \space M\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[A]=3.26 \times 10^{-5} M,\) calculate the value of \(k\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(\mathrm{B}\) and the concentration of A after 10.0 min.
For the reaction \(A+B \rightarrow C\), explain at least two ways in which the rate law could be zero order in chemical A.
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.
Hydrogen peroxide and the iodide ion react in acidic solution as follows: $$\mathrm{H}_{2} \mathrm{O}_{2}(a q)+3 \mathrm{I}^{-}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ The kinetics of this reaction were studied by following the decay of the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) and constructing plots of \(\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\) versus time. All the plots were linear and all solutions had \(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_{0}=8.0 \times 10^{-4} \mathrm{mol} / \mathrm{L} .\) The slopes of these straight lines depended on the initial concentrations of \(\mathrm{I}^{-}\) and \(\mathrm{H}^{+} .\) The results follow: The rate law for this reaction has the form $$\text { Rate }=\frac{-\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}=\left(k_{1}+k_{2}\left[\mathrm{H}^{+}\right]\right)\left[\mathrm{I}^{-}\right]^{m}\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]^{n}$$ a. Specify the order of this reaction with respect to \(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\) and \(\left[\mathrm{I}^{-}\right]\) b. Calculate the values of the rate constants, \(k_{1}\) and \(k_{2}\) c. What reason could there be for the two-term dependence of the rate on \(\left[\mathrm{H}^{+}\right] ?\)
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