Question

For a first-order irreversible reaction carried out in a batch reactor, the rate of reaction is proportional to:

• A. Square of concentration
• B. Concentration of reactant
• C. Inverse of concentration
• D. Product concentration

Hint:

Remember the units of the rate constant \( k \) to identify the order:
- Zero order: \( \text{mol}/(\text{L} \cdot \text{s}) \)
- First order: \( \text{s}^{-1} \)
- Second order: \( \text{L}/(\text{mol} \cdot \text{s}) \)
The unit for a reaction of order \( n \) is \( (\text{conc})^{1-n} \text{time}^{-1} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In chemical kinetics, the "order" of a reaction defines the power dependency of the reaction rate on the concentration of the reactants.
A first-order reaction is a chemical reaction in which the rate of reaction is linearly dependent on the concentration of only one reactant.
Step 2: Key Formula or Approach:
The rate law for a general reaction \( A \rightarrow \text{Product} \) is given by:
\[ -r_A = k C_A^n \]
Where:
\( -r_A \) is the rate of reaction.
\( k \) is the rate constant.
\( C_A \) is the concentration of reactant \( A \).
\( n \) is the order of the reaction.
Step 3: Detailed Explanation:
For a first-order reaction, the value of \( n \) is equal to 1.
Substituting this into the rate law equation:
\[ -r_A = k C_A^1 = k C_A \]
This equation clearly shows that the rate of reaction \( (-r_A) \) is directly proportional to the concentration of the reactant \( (C_A) \).
If the concentration is doubled, the rate also doubles; if the concentration is halved, the rate is halved.
Step 4: Final Answer:
Therefore, the rate of a first-order irreversible reaction is proportional to the concentration of the reactant.