Question

If the reaction is \( 3^{\text{rd}} \) order, if the concentration of reactant is doubled, then the rate is

• A. Increased twice
• B. Increased 8 times
• C. Unchanged
• D. Decreased 2 times

Hint:

For these types of questions, simply raise the factor by which concentration changes to the power of the order. If concentration changes by factor $x$, and order is $n$, new rate changes by factor $x^n$. Here, $2^3 = 8$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The order of a reaction dictates how sensitive the reaction rate is to changes in the concentration of the reactants. It is represented by the exponent in the rate law equation.

Step 2: Key Formula or Approach:
Write the generic rate law for a 3rd order reaction involving a single reactant. Then, substitute the new concentration to see how the overall rate changes.

Step 3: Detailed Explanation:
Let the initial concentration of the reactant be $[A]$.
For a reaction that is overall 3rd order with respect to a single reactant, the rate law is:
$\text{Rate}_{initial} = k[A]^3$
The problem states that the concentration of the reactant is doubled.
Let the new concentration be $[A'] = 2[A]$.
Now, calculate the new rate using this new concentration:
$\text{Rate}_{new} = k[A']^3$
Substitute $[A'] = 2[A]$ into the equation:
$\text{Rate}_{new} = k(2[A])^3$
$\text{Rate}_{new} = k \cdot (2^3) \cdot [A]^3$
$\text{Rate}_{new} = k \cdot 8 \cdot [A]^3$
$\text{Rate}_{new} = 8 \cdot (k[A]^3)$
Since $(k[A]^3)$ is the initial rate, we have:
$\text{Rate}_{new} = 8 \times \text{Rate}_{initial}$
Therefore, the rate is increased 8 times.

Step 4: Final Answer:
The rate is increased 8 times.