Problem 58 - A Complete Resource Book in Chemistry for JEE Main

Question

\(k\) for a zero-order reaction is \(2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). If the concentration of the reactant after \(25 \mathrm{~s}\) is \(0.5 \mathrm{M}\), the initial concentration must have been (a) \(0.5 \mathrm{M}\) (b) \(1.25 \mathrm{M}\) (c) \(12.5 \mathrm{M}\) (d) \(1.0 \mathrm{M}\)

Step-by-Step Solution

Answer

The initial concentration was 1.0 M (option d).

Step 1: Understand the zero-order reaction equation

In a zero-order reaction, the rate does not depend on the concentration of the reactants. The formula is given by \([A]=[A_0]-kt\), where \([A]\) is the concentration after time \(t\), \([A_0]\) is the initial concentration, and \(k\) is the rate constant.

Step 2: Substitute the given values

We know: \([A] = 0.5 \mathrm{~M}\), \(t = 25 \mathrm{~s}\), and \(k = 2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Substitute these into the equation: \[0.5 = [A_0] - (2 \times 10^{-2}) \times 25\]

Step 3: Calculate the change in concentration

Calculate the term \(kt\) to find out how much concentration has changed over the time interval. \(kt = (2 \times 10^{-2}) \times 25 = 0.5\) This means the concentration has decreased by 0.5 M.

Step 4: Solve for initial concentration

Using the equation from Step 2, solve for \([A_0]\): \[0.5 = [A_0] - 0.5\]Rearrange the equation to find \([A_0]\): \[[A_0] = 0.5 + 0.5 = 1.0 \mathrm{M}\]

Step 5: Verify the solution

Compare the calculated initial concentration \([A_0] = 1.0 \mathrm{M}\) with the options provided. Option \(d\) corresponds to the value we calculated, confirming it is the correct solution.

Key Concepts

Rate Constant

In chemical kinetics, the rate constant, often denoted by the symbol "\( k \)," is a pivotal factor that helps determine how fast a reaction proceeds. For a zero-order reaction, the rate constant has units of concentration per time, typically \( \text{mol L}^{-1} \text{s}^{-1} \). This signifies how much of the reactant concentration changes every second. The specific nature of the rate constant makes it unique to the type of reaction it is associated with. In the case of our exercise, the rate constant is \( 2 \times 10^{-2} \text{ mol L}^{-1} \text{s}^{-1} \). This implies that every second, there is a decrease of \( 2 \times 10^{-2} \text{ mol L}^{-1} \) in the concentration of reactants, regardless of how much of the reactant is left at any given time.
The simplicity of zero-order reactions lies in this constant rate:

• Unlike other reactions where rate may change with concentration, zero-order reactions maintain a constant rate until the reactants are exhausted.
• For practical application, understanding the rate constant helps in predicting how long it will take to reach a desired concentration.

Initial Concentration

Determining the initial concentration \([A_0]\) of a reactant is crucial to understanding the progress of a reaction. In a zero-order reaction, the formula \([A] = [A_0] - kt\) helps us track changes from the starting amount. When calculating \([A_0]\), we rearrange this equation after deriving the concentration at a given time \([A]\) and the interval of time \(t\).
In our example problem:

• The concentration after 25 seconds \([A]\) is found to be 0.5 M.
• The rate constant \(k\) is \(2 \times 10^{-2} \text{ mol L}^{-1} \text{s}^{-1}\).
• The calculation goes as follows: \([A_0] = [A] + kt\). With the known values, this simplifies to \([A_0] = 0.5 + 0.5\), thus \([A_0] = 1.0 \text{ M}\).

Understanding the initial concentration gives a reference point for reaction completion, planning experiments, and adjusting conditions to favor desired chemical processes.

Reaction Rate

The reaction rate in a zero-order reaction is a consistent and straightforward metric. Here, it is invariant with the concentration of the reactants, differentiating it from first-order and second-order reactions where the concentration of reactants influences the rate.
For zero-order kinetics:

• The rate of reaction is always equal to the rate constant \( k \), regardless of the concentration present.
• This means the depletion of reactants occurs at an unwavering pace until the reactants are used up.
• In our specific example, the reaction rate is simply \(2 \times 10^{-2} \text{ mol L}^{-1} \text{s}^{-1}\), meaning every second, \(2 \times 10^{-2} \text{ mol/L}\) of the reactant is consumed.

This constant rate underscores the predictability of zero-order reactions, making them easier to manage in experimental and industrial settings where consistency is paramount.