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My teacher gave us the equations for zero-order, first-order and second-order reactions as follows:

$$ [A]_t = -akt + [A]_0 $$ $$ \ln [A]_t=-akt+\ln [A]_0$$ $$\frac1{[A]_t}=akt+\frac1{[A]_0}$$

Where a is the coefficient of $A$ in the chemical reaction, but everywhere I've seen, the $a$ is not there. I have the Central Science 12th edition and there's no $a$ either, is my teacher right? Because that value does affect the answer.

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From the reaction

$$\ce{2A -> B}$$

we have $a = 2$, which leads to the differential equation:

$$\frac{d[A]}{dt} = - k [A]^2$$

which then leads to the integrated rate law you've cited. The $2$ is nowhere as a coefficient, it is related to the order of the equation, and determines which integral/form should be used. Similar conventions also apply for 1st and zeroth order kinetics. I disagree with your instructor.

It should be stressed that these rate laws apply to elementary reactions, not complex reactions. Experimental measurement may differ from that predicted using an elementary approach, which then leads to the conclusion there are other steps involved. In the case of OP's comment regarding oxidation of carbon monoxide with NO2, elementary reactions would predict a rate proportional to $$k [\ce{NO2}][\ce{CO}]$$ if the reaction is the result of a simple bimolecular collision. The fact that a different rate is observed indicates this process is more elaborate.

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    $\begingroup$ And I must disagree with you because the order of the equation has nothing to do with the coefficient of the reagent. Particular orders must be found according to experimental data... or so I've been taught. $\endgroup$ – ChairOTP Feb 27 '13 at 21:09
  • $\begingroup$ You might want to reexamine that thought in the context of physical chemistry, and not just parrot 'what you've been taught'. For any elementary reaction, the coefficient alone determines the order of reaction. The probability of any two [A] molecules colliding to reaction MUST be proportional to $$\[A]^2$$ (just think about it). For any complex mechanism involving several steps, the apparent rate law may not have such a straightforward relation, but that is beyond the context of OP's question and therefor irrelevant. $\endgroup$ – Lighthart Feb 27 '13 at 21:14
  • $\begingroup$ I quote "For any reaction, the rate law must be determined experimentally. In most rate laws, reaction orders are 0, 1, or 2. However, we also occasionally encounter rate laws in which the reaction order is fractional or even negative." Also "Although the exponents in a rate law are sometimes the same as the coefficients in the balanced equation, this is not necessarily the case." And it's not beyond the context of my question. $\endgroup$ – ChairOTP Feb 27 '13 at 21:21
  • $\begingroup$ Again with the quoting, and not with the thinking. Elementary reactions follow, without exception the example I've shown. Complex reactions do not. An example of an elementary would be the coupling of radicals by combination. An example of a complex rxn would be the acid catalyzed synthesis of ether from ethanol. For radical combination, there is one step, and the math works flawlessly. For acid-catalyzed etherification, there are three steps, and all must be considered, and the overall process may no longer have the same rate as a simple elementary reaction. $\endgroup$ – Lighthart Feb 27 '13 at 21:25
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    $\begingroup$ It really doesn't "really" matter if you have $k$ or $2k$ as long as all parties involved know which you are using. The $k$ can "absorb" a constant term (which then changes the value of $k$...). Wikipedia notes some dis/advantages of this here: en.wikipedia.org/wiki/Rate_equation#Second-order_reactions $\endgroup$ – tyler Feb 27 '13 at 21:58

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