How do you find the rate law, given only the chemical reaction?
Rate law for a reaction.... The question is how do you get the rate law for a reaction when you only have the reaction, and I assume its balanced chemical equation. The answer is, "You don't." Rate laws must be determined experimentally. Unlike equilibrium expressions, the rate law cannot be determined from the chemical equation. Unless..... Unless the equation is KNOWN to be an elementary step. Even though a chemical reaction is given by a single chemical equation, the actual reaction may occur in multiple steps, called the reaction mechanism. Each step in the mechanism is an elementary step, because it cannot be broken down into simpler steps. Together, all of the elementary steps is the mechanism. The rate law is usually based on the slow step (if there is only one slow step), called the rate determining step. A rate law for an elementary step is based on the law of mass action, and the rate equation can be deduced from the coefficients in the balanced equation. But that is only true if we know the rate determining step, and usually that is determined experimentally. In fact, the mechanism is usually conjecture and there may be several possible mechanisms. The bottom line: The rate law for a reaction is determined experimentally, it is not determined from a chemical equation.
Calculating the reaction rate law?
To do these, first look for a set of experiments where only one concentration is changed, exp 1 & 2 fit this requirement. Note that the only difference between 1 & 2 is that [NO] is doubled from 0.001 to 0.002 and the observed rate went up by a factor of 4 (0.002 to 0.008), so the reaction is second order in [NO]. Note that in exp 3, increasing the [NO] by a factor of 3 ups the rate by a factor of 9 (3^2); this confirms that the reaction is second order in [NO]. Now, we need to find the order with respect to [H2]. compare exps 4 & 5, note that increasing the [H2] by a factor of 2 increased the rate by a factor of 2 so it's first order in [H2]. Exp 6 confirms this, as increasing the [H2] by a factor of 3 increased rate by a factor of 3. So reaction is 2nd order in [NO] and first order in [H2]. This is not the only way to solve this from the provided data, but probably the simplest way. To get the value for k, take the observed rate and divide by the [NO]^2 x [H2], (do it for several of the sets of concentrations to make sure there aren't any mistakes) for all cases, you should get the same (or nearly the same) value, in this case it's 500,000 for k. This last part I'm not absolutely certian of (it has been many years since I did these), so check your notes/book. The units for k are what ever it needs to get the final answer to be in units of m/l s. In the case of, [NO]^2 x [H2], the units would be m^3/L^3; we need the units of m/Ls for the final rate values, so k needs the units L^2/m^2 s. k L^2/m^2 s x [m^3/L3] = L^2 x m3/m^2 L^3 s After canceling, goes to m/L s The final rate being (in mol/L s) = 500,000 L^2/M2 s [NO]^2[H2]
Can rate constant of a chemical reaction be negative?
If you are only examining a single reaction, then the answer is no. Given that the pre-exponential factor of the Arrhenius equation is always positive (otherwise, you would potentially have a rate constant of 0, meaning the reaction never occurs), the rate constant is never negative.[math]k = Ae^{\frac{E_{a}}{RT}}[/math]Note that the exponential, e^x, has a range only defined over the non-negative real numbers (and approaches zero as x approaches negative infinity). This means that the only possible way the rate constant could ever be negative is if the pre-exponential factor, A, is negative. There are no experimental cases I know of where A is negative, and a negative A just doesn't make sense. An A of 0 doesn't even make sense, as you would never know that the reaction exists, because it would never occur, despite changing the temperature.However, there is a running convention, especially in the biochemical community, to denote the reverse reaction of an equilibrium's rate constant as being the negative of the forward reaction's rate constant. This is meant to be explanatory, to provide you with an orientation regarding the equilibrium reaction. You can even calculate equilibria by relying on this convention. But if you examine the reverse reaction alone, you are changing your frame of reference, so the rate constant will necessarily become positive.If you were to examine the forward and reverse reactions as if the reverse reaction was the forward reaction, and vice versa, the forward reaction would have a negative rate constant. Since you know two things that contradict each other cannot coincide, there must be an explanation. The explanation is that you are changing frames of reference, thus your observations change correspondingly. If you do not couple the two reactions, and just examine each one in it's respective frame of reference, both of them will have a positive rate constant, and this makes sense kinetically.
Is the rate expression of an elementary reaction only dependent on the stoichiometric ratio of the reactants?
If you define “elementary reactions” as single step, molecular without intermediate then I suppose that is true. Kind of. But not useful.It is just as true (and useful) as saying that in cloudcuckooland all cuckoos live on clouds, because you would be hard pressed to find a truly “elementary reaction” as defined. Most reactions are not that simple and must be investigated in detail under various conditions. Some might look “elementary” under some condition but not under others.The truth is that the exponents in a rate equation can only safely be put equal to the stoichiometric coefficients if the reaction is sufficiently close to equilibrium. Far away from that all bets are off.I think that the invention of the educational concept “elementary reactions” is an example of trying to make the world look like how simple the teacher would like it to be rather than what the world is really like. This is atrociously bad teaching, because it saddles a lot of students with indelible misconceptions.One hour of teaching misconceptions by one teacher requires 10 hours of trying to un-teach later on by another teacher, because un-teaching is lot hard than teaching.A point I should add is that rate expressions (and their exponents) can changeThis happens when you start to approach equilibrium but it also happens when you add a catalyst. If all reactions were truly “elementary reactions” catalysis would not exist. Is that really what we want our students to think?
How do I write a rate law for an overall reaction given a mechanism where the 1st step is rate determining?
If the first step is known to be the rate-determining one, then use the reactants in that equation. rate = k[H2O2][I-] . . .since both reactants have coefficients of 1, then the order for each reactant is 1. That is, rate = k [H2O2]^1 [I-]^1
How can you determine a catalyst for a given reaction?
This is one of the fields of inorganic chemistry. There are many things you need to understand about a catalyst to know what it will do.First, when you state that the “catalyst doesn’t specifically react”, that’s the wrong way to think about it. It does react! That’s what makes it an effective catalyst. However, the other part about being a catalyst is that it needs to be recovered after the reaction, or at least turn over the reaction many, many times before being consumed.For many catalysts, yes, it’s a little bit of trial and error. However, once we understand the mechanism of how these catalysts are involved in the reaction, we can then “tune” them to work better or use them as catalysts for completely different reactions. Many clues are taken from biochemistry. For example, you might find that a particular enzyme from an almond will do a specific reaction that adds a stereo-center to a molecule and results in only one enantiomer. That’s a useful catalyst! If that enzyme was expensive to isolate/produce, one might try to build analogs using pure organometallic synthetic methods.In any case, take an inorganic chemistry class. The basics you learn there will introduce new ideas that you may not have gotten from organic chemistry. The d-orbitals in transition metals are fascinating things. For example, you’ll learn about how carbon monoxide can bond to a metal and one of that metal’s electron-filled d-orbitals overlaps with carbon monoxide’s anti-bonding pi molecular orbital, thus weakening the C-O bond. If that bond is weakened, it is more susceptible to being broken in another reaction.
How do I find order and molecularity of a chemical reaction? What is the difference between them?
The molecularity of a reaction simply refers to the number of molecules involved in the rate limiting step of that reaction. The rate limiting step is the slowest step to take place in the intermediate pathway between the reactants and final products. For example, suppose compounds A and B react to form compound C. The reaction does not necessarily take place in one step. It may be necessary for compound A to first ionize before it can react with compound B, so that the reaction mechanism is: 1. A => A(-) 2. A(-) + B => C There are 2 major steps in the overall reaction of A + B => C. The molecularity of this reaction depends on which of these is the rate limiting step, or which one has a greater energy of activation. (1) If the first step, ionization of A, is rate-limiting (ie it is the slowest of the 2 reactions), then the reaction is unimolecular, as there is only one molecule involved. The concentration of B will not have an effect on reaction rate, because it is not involved in the rate limiting step. The equation for the rate of this unimolecular reaction: rate=k[A], where k is an experimentally derived constant for the specific reaction. The reaction is then said to follow first order kinetics. (2) On the other hand, if the second step is rate limiting, then the reaction is considered to be bimolecular. The rate of the reaction depends on the concentrations of both A and B and the reaction rate will be written as: rate=k[A][B]. Again, k is an experimentally derived constant for this specific reaction and will be different if the reaction is of different molecularity. It is possible to determine the molecularity of a reaction experimentally. By carrying out the reaction multiple times and varying the concentrations of the reactants each time, you can find the equation that best represents the reaction. Of course you will have to measure the reaction rate as you go along so that you have the three variables [A], [B], and rate. When you put these variables into one of the previous equations and get the same value for k for each experiment, then you have found the molecularity. The rate of a chemical reaction is the amount of substance reacted or produced per unit time. The rate law is an expression indicating how the rate depends on the concentrations of the reactants and catalysts. The power of the concentration in the rate law expression is called the order with respect to the reactant or catalyst.
Are chemical reactions consistently based on laws of physics?
No, I wouldn’t classify a chemical reaction as a law. There are chemical laws which can help predict whether a given reaction will occur, and with what consequences, in terms of products formed as well as mass and energy accounting.Using reactions analagous to known reactions is sometimes a good start when making predictions. In the case you mentioned, alkali metals have similar valences and other properties that make it a fair bet. But because reactions look similar is no guarantee that they will occur in just the same way, or at the same rate, or even in the same direction.It is true that, at the basic level, chemical processes, like all natural processes, are consistent with the laws of physics. In the case of chemical reactions, however, the applicable physical laws are largely rooted in the field of quantum mechanics. Although all reactions obey these laws perfectly, that doesn’t mean that the QM laws themselves are of practical use in doing chemistry. Trying to predict a chemical reaction by using a quantum-mechanical analysis of all the possible quantum states of every electron in every atom of the reactants and products is a completely impractical way of going about it.The chemical laws chemists use on a daily basis are generally approximations and generalizations of the behavior of materials that are less computationally daunting than particle physics, and do a good enough job to accomplish a useful given task.
Elementary rate law?
A + 5/3 B ---->>> is not an elementary step - you cannot have a fraction of a molecule reacting. So you cannot write a rate law from this. Elementary steps in a reaction mechanism cannot have fractional coefficients. 3A+5B-----> could be elementary - but it is unlikely because molecularities greater than 3 are unheard of. If this is elementary the rate law would be rate = k[A]^3[B]^5 if it is not elementary you could not know the coefficients. So would have to write rate = k[A]^x[B]^y Chances are that neither of the reactions you write are elementary and you still need to look at the reaction mechanism
What is a chemical reaction?
A chemical reaction is a process that leads to the transformation of one set of chemical substances to another.Classically, chemical reactions encompass changes that only involve the positions of electrons in the forming and breaking of chemical bonds between atoms, with no change to the nuclei (no change to the elements present), and can often be described by a chemical equation. Nuclear chemistry is a sub-discipline of chemistry that involves the chemical reactions of unstable and radioactive elements where both electronic and nuclear changes can occur.The substance (or substances) initially involved in a chemical reaction are called reactants or reagents. Chemical reactions are usually characterized by a chemical change, and they yield one or more products, which usually have properties different from the reactants. Reactions often consist of a sequence of individual sub-steps, the so-called elementary reactions, and the information on the precise course of action is part of the reaction mechanism. Chemical reactions are described with chemical equations, which symbolically present the starting materials, end products, and sometimes intermediate products and reaction conditions.Chemical reactions happen at a characteristic reaction rate at a given temperature and chemical concentration. Typically, reaction rates increase with increasing temperature because there is more thermal energy available to reach the activation energy necessary for breaking bonds between atoms.