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Save to playlist. The results are as follows: [A] [B] Rate 0. Explanation : When the concentration of A stays the same but the concentration of B doubles, the rate quadruples, showing a second order rate law based on B.
Possible Answers: always requires experimental data to determine. Correct answer: is measured in unit of concentration over time. Explanation : Rate laws can rarely be determined from just the reaction; they usually require experimental data. Consider the following balanced equation. Possible Answers:. Correct answer:. Explanation : Since the rate law is provided for the reaction, we can plug in the values that are relevant to the rate of the reaction.
Note that HF is not included; the reaction must be zero order with respect to HF. Consider the following reaction and experimental data.
Explanation : When determining the rate law for a reaction, you need to determine the order for each reactant in the reaction. Now that we have the orders for each reactant, we can write the rate law accordingly. Using the above trials, write the rate law for the reaction. Correct answer: I and III only. Explanation : The rate based on concentration is related to the coefficients in front of the compounds.
Correct answer: Half of ICl, same as I2. Explanation : By the stoichiometry, since 1 mole of of H2 and 2 moles of ICl produces 2 moles of HCl and 1 mole of I2 , we know that H2 is consume half as fast as ICl and produced at the same rate as I2. Which of the following is a classic example of a first-order reaction? Possible Answers: A collision between 2 reactant molecules. Correct answer: Radioactive decay. Explanation : First order reactions have rates that are directly proportional to only 1 reactant.
Copyright Notice. View AP Chemistry Tutors. Jenson Certified Tutor. Erik Certified Tutor. Mikhail Certified Tutor. Therefore, if we were to graph the natural logarithm of the concentration of a reactant ln versus time, a reaction that has a first-order rate law will yield a straight line, while a reaction with any other order will not yield a straight line Figure The slope of the straight line corresponds to the negative rate constant, — k , and the y -intercept corresponds to the natural logarithm of the initial concentration.
We are given the rate constant and time and can determine an initial concentration from the number of moles and volume given. We can substitute this data into the integrated rate law of a first-order equation and solve for the concentration after 2. The rate for second-order reactions depends either on two reactants raised to the first power or a single reactant raised to the second power. The differential rate law can be written:. Therefore, if the reaction is second order, a plot of versus t will produce a straight line with a slope that corresponds to the rate constant, k , and a y -intercept that corresponds to the inverse of the initial concentration, Figure Zero-order reaction rates occur when the rate of reactant disappearance is independent of reactant concentrations.
When presented with experimental concentration—time data, we can determine the order by simply plotting the data in different ways to obtain a straight line. The reaction rate will therefore quadruple.
B The rate law tells us that the reaction rate is constant and independent of the N 2 O concentration. That is, the reaction is zeroth order in N 2 O and zeroth order overall. C Because the reaction rate is independent of the N 2 O concentration, doubling the concentration will have no effect on the reaction rate.
B The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall.
C Doubling the initial cyclopropane concentration will increase the reaction rate from k [cyclopropane] 0 to 2 k [cyclopropane] 0. This doubles the reaction rate.
Given the following two reactions and their experimentally determined differential rate laws: determine the units of the rate constant if time is in seconds, determine the reaction order with respect to each reactant, give the overall reaction order, and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled.
The number of fundamentally different mechanisms sets of steps in a reaction is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanisms can simplify what might seem to be a confusing variety of chemical reactions. This can be done by designing experiments that measure the concentration s of one or more reactants or products as a function of time.
To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A.
Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied.
In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments. Because 1. We can cancel like terms to give 0. By selecting two experiments in which the concentration of B is the same, we were able to solve for m.
Conversely, by selecting two experiments in which the concentration of A is the same e. Canceling leaves 1.
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