Problem 38 - A Complete Resource Book in Chemistry for JEE Main

Question

If the rate law of a reaction \(\mathrm{nA} \longrightarrow \mathrm{B}\) is expressed as Rate \(=-\frac{1}{n} \frac{d[A]}{d t}=+\frac{d[B]}{d t}=k[A]^{x}\) The unit of the rate constant will be (a) \(\mathrm{mol}^{\mathrm{x}} / \mathrm{L}^{\mathrm{x}} / \mathrm{s}\) (b) \(\mathrm{L}^{\mathrm{x}} / \mathrm{mol}^{\mathrm{x}} \mathrm{s}\) (c) \(\operatorname{mol}^{(1-x)} / L^{(x-1)} \cdot S^{-1}\) (d) \(m o l^{(x-1)} / L^{(1-x)} \cdot S^{-1}\)

Step-by-Step Solution

Answer

The unit of the rate constant is (c) \( \text{mol}^{1-x} / \text{L}^{x-1} \cdot \text{S}^{-1} \).

Step 1: Understanding the Rate Law

The rate law is given by \[ \text{Rate} = k [A]^x \] where 'Rate' is expressed in terms of concentration change over time, hence its unit is \( \text{mol L}^{-1} \text{s}^{-1} \). Here, \( k \) is the rate constant, \([A]\) is the concentration of reactant \(A\) in terms of \( \text{mol L}^{-1}\), and \(x\) is the order of the reaction with respect to \(A\). The order can be determined experimentally and can be zero, a positive fraction, or a whole number.

Step 2: Deriving the Unit of Rate Constant

We need to express the units of \( k \). Given the rate law equation: \[ \text{Rate} = k [A]^x \]Substitute units, \[ \text{mol L}^{-1} \text{s}^{-1} = k (\text{mol L}^{-1})^x \]Rearranging for \( k \), we get:\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^x} \]

Step 3: Simplifying the Units

To simplify, express the denominator:\[ (\text{mol})^x (\text{L}^{-1})^x = \text{mol}^x \text{L}^{-x} \]Thus, the rate constant \( k \) becomes:\[ k = \text{mol}^{1-x} \text{L}^{x-1} \text{s}^{-1} \]

Step 4: Matching With Provided Options

Comparing the derived unit \( \text{mol}^{1-x} \text{L}^{x-1} \text{s}^{-1} \) with the options:(a) \( \text{mol}^{x} / \text{L}^{x} / \text{s} \) (b) \( \text{L}^{x} / \text{mol}^{x} \text{s} \)(c) \( \text{mol}^{1-x} / \text{L}^{x-1} \cdot \text{S}^{-1} \) (d) \( \text{mol}^{x-1} / \text{L}^{1-x} \cdot \text{S}^{-1} \)The correct match for \( k = \text{mol}^{1-x} \text{L}^{x-1} \text{s}^{-1} \) is option (c).

Key Concepts

Reaction Order

The concept of reaction order is fundamental in understanding how the concentration of reactants influences the rate of a chemical reaction. Reaction order is denoted by the exponent in the rate law equation given by \[ \text{Rate} = k [A]^x \] where - "Rate" is the speed of the reaction,- \([A]\) is the concentration of reactant \(A\),- \(x\) is the reaction order with respect to \(A\).

Reaction order can be:- **Zero**: The rate is independent of the concentration of reactants. - **First**: The rate is directly proportional to the concentration. - **Second**: The rate is proportional to the square of the concentration.

Keep in mind that reaction order is determined experimentally and can even take on fractional values. Understanding the reaction order helps chemists predict how the rate of reaction changes as concentrations change.

Rate Constant

The rate constant, \(k\), is a crucial part of the rate law equation, which determines the reaction's speed. It connects the rate of reaction to the concentrations of the reactants raised to their respective powers. For a given reaction like \( \mathrm{nA} \longrightarrow \mathrm{B} \), the rate law \[ \text{Rate} = k [A]^x \] includes the rate constant \(k\) which is specific to the reaction at a given temperature.

A few points about rate constant:- **Independent of concentration**: Unlike reaction rate, \(k\) does not change with the concentration changes.- **Temperature-dependent**: The rate constant changes with temperature; usually increases with temperature.- **Unit-dependent on reaction order**: The units of \(k\) will differ based on the order of the reaction. This is because it needs to compensate for the units of concentration to ensure overall dimensions balance with the rate.

Knowing the rate constant allows prediction of how fast a reaction will occur under specified conditions.

Units of Rate Constant

The units of the rate constant are vital to ensure that the rate law equation is dimensionally consistent. For an equation like \[ \text{Rate} = k [A]^x \] the units of the rate are generally \( \text{mol L}^{-1} \text{s}^{-1} \).

To find the units of the rate constant \(k\), equate the units of rate to the product of \(k\) and the concentration of reactants. Let's simplify:

• Start with the basic form: \( k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^x} \)
• Simplify: \( k = \text{mol}^{1-x} \text{L}^{x-1} \text{s}^{-1} \)

The units depend heavily on \(x\), the reaction order:- **Zero-order**: \( \text{mol L}^{-1} \text{s}^{-1} \)- **First-order**: \( \text{s}^{-1} \)- **Second-order**: \( \text{L mol}^{-1} \text{s}^{-1} \)

By understanding the units of \(k\), students can reinforce their understanding of both the rate of reaction and its dependency on reactant concentrations.