The conversion of molecules A to B follows second order kinetics. If the concentration of A is increased to four times, how will the rate of formation of B be affected?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Increases by 16 times.
In a second-order reaction, the rate law is given by:
\(\text{Rate} = k [A]^2\)
Here, the rate depends on the square of the concentration of reactant \( A \).
Understanding the problem:
If the concentration of \( A \) is increased to four times its original concentration, we can express this as:
\([A_{\text{new}}] = 4 \times [A_{\text{initial}}]\)
Effect on rate:
Substituting the new concentration of \( A \) into the rate law:
\(\text{Rate}_{\text{new}} = k (4 \times [A_{\text{initial}}])^2 = k (16 \times [A_{\text{initial}}]^2)\)
Thus, the new rate is
\(\text{Rate}_{\text{new}} = 16 \times \text{Rate}_{\text{initial}}\)
Conclusion:
The rate of formation of \( B \) increases by 16 times when the concentration of \( A \) is increased to four times. Therefore, the correct answer is option (3): Increases by 16 times.