Question

In the chemical reaction A $\rightarrow$ B, what is the order of the reaction? Given that, the rate of reaction doubles if the concentration of A is increased four times.

• A. 2
• B. 1.5
• C. 0.5
• D. 1

Hint:

The order of a reaction can be determined experimentally using the given rate conditions by applying logarithmic methods to compare relative rate changes.

Answer 1

The Correct Option is C

Step 1: Understanding the Rate Law 
The rate of a reaction is given by: \[ r = k[A]^n \] where \( k \) is the rate constant, \( [A] \) is the concentration of reactant, and \( n \) is the order of the reaction. 
Step 2: Applying Given Conditions 
If the concentration of \( A \) is increased four times, the rate doubles. Mathematically, \[ 2r_1 = k[4A]^n \] Using the original rate equation: \[ r_1 = k[A]^n \] 
Step 3: Dividing the Equations 
\[ \frac{2r_1}{r_1} = \frac{k(4[A])^n}{k[A]^n} \] \[ 2 = 4^n \] 
Step 4: Solving for \( n \) 
Taking logarithm on both sides, \[ \log 2 = n \log 4 \] \[ \log 2 = n (2 \log 2) \] \[ n = \frac{\log 2}{2 \log 2} = \frac{1}{2} = 0.5 \] 
Final Answer: The order of reaction is 0.5.

Answer 2

Step 1: Understand the relation between rate and concentration
The rate of reaction generally follows the expression:
rate = k [A]^n, where n is the order of the reaction with respect to A.

Step 2: Analyze the given data
When the concentration of A is increased 4 times, the rate doubles.
So, rate₂ / rate₁ = 2 and [A]₂ / [A]₁ = 4.

Step 3: Use the rate relation to find order n
(rate₂ / rate₁) = ([A]₂ / [A]₁)^n
2 = 4^n
Take logarithm on both sides:
log(2) = n × log(4)
n = log(2) / log(4) = 0.3010 / 0.6021 ≈ 0.5

Step 4: Final Conclusion
Therefore, the order of the reaction with respect to A is 0.5.