How does initial rate of reaction imply rate of reaction at any time?

Question

My textbook begins the Chemical Kinetics section by describing the "Initial Rates Method" of determining the rate of reaction. I understood it as the following (for a first order reaction): $$\ce{A -> B+C}$$ $$\text{rate}_\mathrm{i}=k[\ce{A}]_\mathrm{i}$$

The subscript $\text{i}$ shows that it is the initial rate. This I agree with, as it is suggested by experimental data. However, on the next page, the textbook makes a leap and (implicitly) states that this relationship holds at any time: $$\text{rate}=-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=k[\ce{A}]$$ This becomes a differential equation, the justification for which the textbook did not provide. How do we know that the change of concentration with respect to time depends on the concentration itself? I don't see how this can be assumed from the initial rates method.

Answer 1

First things first

When we talk about the rate of a reaction, we mean the rate at which the products appear. This means that it is also equal to the disappearance of the reactants. So, when we quantify the rate in terms of reactants, they must be in opposing magnitude. $$\text{rate }= \frac {\Delta [\text{product}]}{\Delta t}=-\frac{\Delta [\text{reactant}]}{\Delta t}$$

From A to B (and C)

Suppose that I am running the reaction $\ce{A ->B +C}$ in two separate vessels, flask 1 and flask 2, each with a different concentration of $\ce{A}$. We'll say that flask 1 is more concentrated. If you agree that the initial rate is given by $\text{rate}_\mathrm{i} = k[\ce{A}]_\mathrm{i}$, then it follows that for each the initial rate is $\text{rate}_{i,fl1} = k[\ce{A}]_{i, fl1}$ and $\text{rate}_{i,fl2} = k[\ce{A}]_{i, fl2},$ respectively.

If we were to add something to flask 2, would it change the reaction rate? As long as we didn't add any $\ce{A},$ the rate would be the same. The rate is dependent on $\ce{A}$ but nothing else. So let's add a little bit of $\ce{B}$ and a little bit oc $\ce{C}$. We'll add the same amount of both. Let's call that amount $x$ and set it equal to the difference in concentrations in $\ce{A}$ in the two flasks. $$x=[\ce{A}]_{i,f1}-[\ce{A}]_{i.f2}$$

Now let's let the reaction in flask 1 proceed until the concentration of $\ce{A}$ is equal to the amount in flask 2. To distinguish the conditions now from the initial conditions, we'll call this time $t=1$ (arbitrary units). By the stoichiometry of the reaction, we know the amount of $\ce{A}$ and $\ce{A}$ will be $\Delta[\ce{A}] = [\ce{A}]_{i,f1}-[\ce{A}]_{i,f2}=x$ There is no perceivable difference between the two flasks, and the reaction rate is thus the same in flask 1 as in flask 2. Because we have chosen arbitrary concentrations, we can abstract it to any concentration. $$\text{rate}_{t=1,f1}=\text{rate}_{i,f2}=k[a\ce{A}]_{i,f2}=k[\ce{A}]$$

Finally, we want the instant rate. We do this by decreasing the time interval to an infinitesimal amount. $$\lim_{\Delta t \to 0}\frac{-\Delta[\ce{A}]}{\Delta t}=-\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t}=k[\ce{A}]$$

Answer 2

A first order reaction is defined as a reaction in which the reaction rate obeys the equation $$r(t) \equiv \mathrm{d}[A(t)] / \mathrm{d}t = k[A(t)] \tag{1}$$ Your first equation $$r(0) \equiv \mathrm{d}[A(0)] / \mathrm{d}t = k[A(0)] \tag{2}$$

is just a particular situation. We cannot know a priori if a reaction that obeys (2) will obey (1). We can only measure how $r(t)$ varies over time. If $r(t)$ follows equation (1) then it is said that the kinetics follow a first-order equation.

In chemical reactions of the type $\ce{A -> B + C}$ it is natural to expect a first order reaction at initial times. As long as the reaction doesn't need any other molecule or agent in the medium (aside from perhaps other essentially constant factors) than $\ce{A}$, and doesn't involve collisions between two or more molecules to ignite the reaction, one can expect the rate to be proportional to the number of molecules present at each time, hence to $[\ce{A}(t)]$.

After a long time, when the reactive concentration is depleted and the concentration of the products builds in (assuming no diffusion of the products) the first order law is typically violated, since the back reaction $\ce{B + C -> A}$ (aside from other collisions) can occur. In particular, is is not possible to reach equilibrium conditions under the first order law.