# Rate Law

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In a chemical reaction,  it is important to consider not only the chemical properties of the reactants, but also the conditions under which the reaction occurs, the mechanism through which it takes place, the rate at which it occurs, and the equilibrium toward which it is proceeding. The substances which influence the rate of reaction are usually one or more of the reactant’s side, but sometimes it includes products also. Catalysts, which do not appear in the balanced chemical equation, can also influence the rate of reaction.

The rate law of a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentrations of the reactants participating in it.

Example:

If a reaction is given by $aA + bB \to cC + dD$

Where a, b, c, and d denotes the stoichiometric coefficients of the reactants and products, the rate equation for the reaction is given by:

$Rate \propto [A]^x[B]^y$

$\Rightarrow Rate = k[A]^x[B]^y$

1. $[A]$ & $[B]$ denote the concentrations of the reactant side $A$ & $B$.

2. The proportionality constant $‘k’$ is the rate constant for the reaction.

3. $x$ & $y$ denote the partial reaction orders for the reactant side $A$ & $B$ (This may or may not be equal to their stoichiometric coefficients $a$ & $b$).

Note: The expression of the rate law for a specific reaction can only be determined experimentally. The rate law expression is not determined by the balanced chemical equation.

## Order of Reaction:

If Rate is given by $k[A]^x[B]^y$, the overall order of the reaction $(n) = x+y$.

The order of a reaction is the sum of the powers of the concentrations of the reactants in the rate law expression. In the above general reaction, $x$ and $y$ are the powers. Their sum will give the order of the reaction. The order of a reaction can be $1, 2, 3, 0$ and even a fraction.

The order of a reaction provides information about the change in the rate of the reaction that can be expected by increasing the concentration on the reactant side.

• For the first-order reaction, doubling the reactant concentration will double the reaction rate.

• For the second-order reaction, doubling the concentration of the reactants will quadruple the overall reaction rate.

• In third-order reactions, the overall rate increases by eight times when the reactant concentration is doubled.

• If the reaction is of zero-order, doubling the reactant concentration will have no effect on the reaction rate. A zero-order reaction means that the rate of the reaction is independent of the concentration of the reactants.

### Rate Constant

Since rate constant is given by Rate $= k[A]^x[B]^y$

So, $k = \dfrac{Rate}{[A]^x[B]^y}$

The units are calculated by the following equation: $k = (M \cdot s^{-1}) \times (M^{-n}) = M^{(1-n)} \cdot s^{-1}$.

The concentration is represented in $\text{mol} L^{-1}$ or $M$ and time is represented in $\text{seconds}$.

## Units of Rate Constant:

 Order of Reaction Units of Rate Constant n $M^{1-n} \cdot s^{-1}$ 1 $s^{-1}$ 2 $M^{-1} \cdot s^{-1}$ 0 $M \cdot s^{-1}$

Differential Rate Equations

It used to express the rate of a reaction in terms of the change in reactant concentrations $(d[R])$ with respect to time $(dt)$. So the differential form of the rate expression is given by:

$\dfrac{-d[R]}{dt} = k[A]^x[B]^y$

This equation is used to calculate the instantaneous rate of a reaction, which is the reaction rate under a very small interval time.

Note: Ordinary rate law is a differential rate equation because it shows the instantaneous rate of the reaction.

### Integrated Rate Equation

These equations express the concentration of the reactants in a chemical reaction as a function of time,

such rate equations can be used to check how long it would take for a given percentage of the reactants to be consumed in a chemical reaction, and reactions of different orders have different integrated rate equations.

Integrated Rate Equation for Zero-Order Reactions

For Zero-order reactions

Rate ∝ [R]⁰

⇒ Rate = -d[R]/dt = k[R]⁰ = k х 1

⇒ Rate = -d[R]/dt = k

⇒d[R] = -kdt

Integrating both sides, $[R] = -kt + I$

$[R]_0 = -k × 0 + I = I$

Substituting this value of $I$ in the equation, we get

$kt = [R_0] – [R]$

$k = \dfrac{([R_0] – [R])}{t}$

Where, $[R_0]$ is the initial concentration $( t = 0)$

$[R]$ is the concentration of the reactant at any time $‘t’$

$k$ is the rate constant

Integrated Rate Equation for First-Order Reactions

For first-order reactions

kt = 2.303 log([R₀]/[R])

k = (2.303/t) log([R₀]/[R])

Integrated Rate Equation for Second-Order Reactions

For second-order reactions

$kt = \left(\dfrac{1}{[R]}\right) – \left(\dfrac{1}{[R_0]}\right)$

### Did you know?

The first quantitative study in chemical kinetics was done by German scientist Ludwig Ferdinand Wilhelmy who used polarimetry to investigate the acid-catalyzed conversion of sucrose. However, the English chemist Augustus George Vernon Harcourt is considered to be the first scientist who made a significant contribution to the field of chemical kinetics.

Question: What is the Zero-order Reaction? Explain with a Suitable Example.

Answer: Reactions in which the concentration of the reactants do not change with respect to time and the concentration rates remain constant throughout are called zero-order reactions.

Example: H₂ + Cl₂ → 2HCl

Rate = k[H₂]⁰ [Cl₂]⁰

Rate = k

Question: The Order of a Reaction: (Single option correct)

a) can never be a fraction

b) can never be zero

c) must be a whole number

d) can be a whole number, zero, or a fraction

Answer: Order of reaction depends on the rate of reaction on the reactant side. If the rate is independent of the reactants, then the order of the reaction is zero.  Therefore, the rate law of a zero-order reaction would be Rate ∝ [R]⁰, So the correct option is D.

Question: Give an Example of a Third-order Reaction.