Rate of a reaction changes from 2.48 × 10⁻³ mol⁻¹ sec⁻¹ to 4.96 × 10⁻³ mol⁻¹ sec⁻¹ when concentration of reactant is changed from 0.6 M to 2.4 M respectively, the order of reaction is:
0.5
The order of a chemical reaction describes how the rate of the reaction depends on the concentration of the reactants. For a simple reaction involving a single reactant A, the rate law is often expressed as:
$\text{Rate} = k[\text{A}]^n$
where:
In this problem, we are given two different reaction rates at two different concentrations of the reactant. We can use this information to determine the value of $n$, the order of the reaction.
Let $R_1$ be the rate at concentration $C_1$, and $R_2$ be the rate at concentration $C_2$. According to the rate law:
$R_1 = k[C_1]^n$
$R_2 = k[C_2]^n$
We are given:
Substituting these values into the rate law equations:
Equation 1: $2.48 \times 10^{-3} = k[0.6]^n$
Equation 2: $4.96 \times 10^{-3} = k[2.4]^n$
To find the order $n$, we can divide Equation 2 by Equation 1. This cancels out the rate constant $k$:
$\frac{R_2}{R_1} = \frac{k[C_2]^n}{k[C_1]^n}$
$\frac{4.96 \times 10^{-3}}{2.48 \times 10^{-3}} = \frac{[2.4]^n}{[0.6]^n}$
The left side simplifies to:
$\frac{4.96 \times 10^{-3}}{2.48 \times 10^{-3}} = \frac{4.96}{2.48} = 2$
The right side can be written as:
$\frac{[2.4]^n}{[0.6]^n} = \left(\frac{2.4}{0.6}\right)^n = (4)^n$
So, we have the equation:
$2 = 4^n$
To solve for $n$, we need to find what power $n$ applied to 4 gives 2. We know that the square root of 4 is 2, and the square root is equivalent to raising to the power of 0.5 (or 1/2).
$\sqrt{4} = 4^{1/2} = 2$
Comparing this to $2 = 4^n$, we see that $n = 1/2 = 0.5$.
Therefore, the order of the reaction is 0.5.
| Parameter | Value 1 | Value 2 |
|---|---|---|
| Rate (R) | $2.48 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$ | $4.96 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$ |
| Concentration (C) | 0.6 M | 2.4 M |
| Rate Law Ratio | $\frac{R_2}{R_1} = \left(\frac{C_2}{C_1}\right)^n$ | |
| Substitution | $\frac{4.96 \times 10^{-3}}{2.48 \times 10^{-3}} = \left(\frac{2.4}{0.6}\right)^n$ | |
| Simplification | $2 = (4)^n$ | |
| Solving for n | $n = 0.5$ | |
Based on the calculation using the changes in reaction rate with respect to changes in reactant concentration, the order of the reaction is found to be 0.5.
| Concept | Description | Key Relationship |
|---|---|---|
| Rate of Reaction | Speed at which reactants are consumed or products are formed. | Change in concentration over time. |
| Rate Law | An equation relating the rate of a reaction to the concentrations of reactants. | Rate = $k[\text{A}]^n[\text{B}]^m...$ |
| Rate Constant ($k$) | Proportionality constant in the rate law, specific for a reaction at a given temperature. | Independent of concentration. |
| Order of Reaction ($n, m...$ Total Order $n+m...$) | Experimentally determined exponents in the rate law; indicate sensitivity of rate to concentration changes. | Rate $\propto [\text{A}]^n$ |
The method used in this problem is a common way to determine the order of a reaction when kinetic data is available at different concentrations. Here are some points about reaction order:
Second most abundant element in alloy misch metal is:
Match List-I with List-II:
| List-I | List-II |
|---|---|
| (A) Gel | (I) Hair cream |
| (B) Foam | (II) Dust |
| (C) Emulsion | (III) Cheese |
| (D) Aerosol | (IV) Whipped cream |
Choose the correct answer from the options given below:
Degree of dissociation, when molar conductivity of X at its concentration C is 24.14 and its limiting molar conductivity is 48.28 will be:
A divalent ion of 'V' (Atomic no. 23) in aqueous solution is:
Which of the following sols are correctly matched with their corresponding charges?
(A) Cr(OH)₃ sol : negatively charged sol
(B) TiO₂ sol : positively charged sol
(C) CdS sol : positively charged sol
(D) Gum : negatively charged sol
(E) Silver sol : positively charged sol
Choose the correct answer from the options given below: