# Find the velocity constant K given the speed

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

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.
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 $k$.