Q.A first order reaction is found to have a rate constant, $k = 5.5\times10^{-14}\ \text{s}^{-1}$. Find the half-life of the reaction.

Rate = (change in concentration)/(time), so its unit is mol L^-1 s^-1 (or mol dm^-3 s^-1) — the statement is correct.

Rate of a chemical reaction = -(1/stoichiometric coefficient) x d[reactant]/dt = +(1/stoichiometric coefficient) x d[product]/dt. Since concentration is expressed in mol L^-1 and time in seconds, the rate of reaction has units of mol L^-1 s^-1 (equivalently, mol dm^-3 s^-1).

Average rate is a rate measured over a finite time span; as that span shrinks to zero it becomes the instantaneous rate — so the two are equal in the limit $\Delta t \to 0$.

The average rate of a reaction over an interval $\Delta t = t_2 - t_1$ is:

$\text{Average rate} = -\dfrac{\Delta[R]}{\Delta t} = -\dfrac{[R]_2-[R]_1}{t_2-t_1}$

This is the mean rate over that whole time span and can mask how quickly the rate is actually changing within the interval.

The instantaneous rate at a specific time $t$ is the slope of the concentration-vs-time curve at that exact point:

$\text{Instantaneous rate} = -\dfrac{d[R]}{dt}$

Mathematically, the instantaneous rate is defined as the limit of the average rate as the time interval shrinks to zero:

$\text{Instantaneous rate} = \lim_{\Delta t \to 0} \left(-\dfrac{\Delta[R]}{\Delta t}\right) = -\dfrac{d[R]}{dt}$

So, when $\Delta t$ is taken to be extremely small (in the limiting sense, $\Delta t \to 0$), the chord joining the two points on the concentration–time graph becomes the tangent at a single point — and the average rate numerically coincides with the instantaneous rate at that instant.

The rate 'at a particular moment' (not averaged over an interval) is called the instantaneous rate.

Average rate is calculated over a finite time interval (Delta[R]/Delta t) and changes depending on which interval is chosen. To get the rate AT one specific instant, we let the time interval shrink to zero:

Instantaneous rate = -d[R]/dt (or +d[P]/dt), obtained graphically as the slope of the tangent drawn to the concentration-vs-time curve at that particular time.

Rate depends on reactant concentration; as reactants are consumed, concentration and hence rate fall with time.

For a typical reaction, Rate = k[reactant]^n. As the reaction proceeds the reactant concentration continuously decreases, so the rate also decreases with time (it is maximum at the start and approaches zero near completion).

This is why the initial rate is the highest for most reactions.

For a reaction $3A \to 2B$, the rate of the reaction expressed via each species must be divided by its own stoichiometric coefficient, so rates in terms of A and B are related by the ratio of those coefficients.

Setting up the rate expression: For $3A \to 2B$, the overall rate of reaction is defined uniquely (independent of which species you track) as:

$\text{Rate}=-\frac{1}{3}\frac{d[A]}{dt}=+\frac{1}{2}\frac{d[B]}{dt}$

Solving for d[B]/dt in terms of d[A]/dt:

Given the rate of production of B is $+\dfrac{d[B]}{dt}$, equate:

$\frac12\frac{d[B]}{dt}=-\frac13\frac{d[A]}{dt}$

$\frac{d[B]}{dt}=-\frac{2}{3}\frac{d[A]}{dt}$

(Since A is being consumed, $d[A]/dt$ is itself negative, so $-\frac23\frac{d[A]}{dt}$ works out to a positive quantity, consistent with B being produced.)

For gaseous reactions, concentration is expressed as partial pressure, so rate has units of pressure/time.

Rate of reaction = (change in concentration)/(time). For reactions involving gases, concentration is conveniently measured as partial pressure (atm) instead of mol L⁻¹. Hence rate = Δ(pressure)/Δ(time), giving units of atm s⁻¹.

Average rate is measured over a finite $\Delta t$; as that interval is shrunk toward zero, the average rate converges exactly to the instantaneous rate at that moment.

The average rate of a reaction over an interval is $-\dfrac{\Delta[R]}{\Delta t}$ (or $+\dfrac{\Delta[P]}{\Delta t}$), computed using the change in concentration over a finite time interval $\Delta t$. The instantaneous rate is the rate at one particular instant, given by the derivative $-\dfrac{d[R]}{dt}$.

As the time interval $\Delta t$ used for the average is made smaller and smaller (i.e. $\Delta t \to 0$), the average rate over that shrinking interval approaches the slope of the tangent to the concentration-vs-time curve at that point — which is exactly the instantaneous rate. So the two coincide in the limit $\Delta t \to 0$.

Instantaneous rate is the rate at one exact moment, found as $-d[R]/dt$ — the slope of the tangent drawn to the concentration-time plot at that instant (Δt → 0).

Whereas the average rate is measured over a finite time interval, the instantaneous rate is the rate of the reaction at one particular instant of time. It is obtained mathematically by making the time interval Δt infinitesimally small (Δt → 0), turning the average-rate expression into a derivative:

$\text{Instantaneous rate} = \lim_{\Delta t \to 0} \left(-\frac{\Delta[R]}{\Delta t}\right) = -\frac{d[R]}{dt} = \frac{d[P]}{dt}$

Graphically, if concentration is plotted against time, the instantaneous rate at any chosen time t is given by the slope of the tangent drawn to the curve at that point.

Average rate = (change in concentration) / (time interval) over a finite interval $\Delta t$.

For a reaction R → P, the average rate over the time interval $t_1$ to $t_2$ is

$\text{Average rate} = -\frac{[R]_2-[R]_1}{t_2-t_1} = \frac{[P]_2-[P]_1}{t_2-t_1} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t}.$

A negative sign is used for reactants because their concentration decreases with time, so that the rate (a positive quantity) is obtained. It represents the mean rate over the chosen interval, not the rate at a particular instant (which is the instantaneous rate, obtained by shrinking $\Delta t \to 0$).