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This reaction is of first order in [A]. Consider that

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What is the velocity constant $K$ here?


So let's see... The velocity equation is

$$V = K \cdot [A]^n\cdot[B]^m$$

Assuming that it is bimolecular... But... we don't know if it is bimolecular! All they tell me is that there is some reaction with a molecule $A$ involved.

Anyway, they also tell me that the reaction is of first order... on [A]. It is my understanding that this is called a partial order, and the global order is not necessarily the same, as it would depend on the involvement of any other molecule (which we don't know about). Therefore, it is not necessarily true that the equation of velocity is

$$V = K \cdot [A]^n\cdot[B]^m$$

So how do I even approach this kind of problem?

I can't even calculate the general velocity of the reaction, because I don't know the coefficients of the reagents. Heck, I don't even know how many reagents and products are there!

(I know this is an extremely simple exercise and the answer is also quite basic, but I just fail to find an example of exactly this scenario anywhere).

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You are really confusing yourself. The problem is very simple. First of all, remove the fact that the rate of reaction depends on all reactants from your mind. It is not the case. Sometimes, even though multiple reactants are involved, only some of them actually matter when we need to find the rate. Let me explain it without chemistry.

Consider a factory where a car is made. The first level makes the engine in 1 hour. The second level makes the tyres in 1 hour. The third level makes the rest of the parts in 1 hour. The fourth level puts them all together in 1 day. Now, even though all the 4 levels are involved in the process, it is the slowest level (level IV) that determines the rate of the process because the rest of the levels can keep sending the parts, but the finished car won't exit the factory unless level IV puts them all together. This is the case with chemical reactions too! The slowest step decides the order and rate of the reaction.

Now, back to the problem. Since they have given that the reaction follows first order kinetics in $[A]$, the rate law can be written as $R=k[A]$. After a little bit of integration and other stuff, we have the equation, $k=\frac{ln(2)}{t _{1/2}}$ , where $t _{1/2}$ is the half-life, i.e. time taken for concentration (actually, it's no. of molecules) of reactant to become half its original value. In this problem, $t _{1/2}$=10.0 seconds. Substitute and find the answer.

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It's a first order reaction in $\ce{[A]}$: $$-\frac{d\ce{[A]}}{dt}=k\ce{[A]}$$ By integrating the above equation, we get: $$\ln\frac{\ce{[A_0]}}{\ce{[A]}}=kt$$ Where $\ce{[A_0]}$ is the concentration of $\ce{[A]}$ at $t=0$

So, you plot $(\ln\frac{\ce{[A_0]}}{\ce{[A]}})$ as function of $t$. You get a line which passes through the origin, with a slope equals to $k$.

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Another good approach is to use the method I described here: Finding [A] when looking for the velocity constant Try it and see what you think. It doesn't give the exact answer, but, for this set of data, it gives an excellent approximation.

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