Question

A reaction is of first order with respect to a reactant. How is the rate of reaction affected, if the concentration of the reactant is doubled?

Hint:

In first-order kinetics, rate is directly proportional to concentration. Doubling concentration doubles the rate, tripling it makes the rate three times, and so on.

Answer 1

Step 1: General rate law for first-order reaction.
\[ Rate = k[A]^1 \] Step 2: Effect of doubling concentration.
If concentration of \([A]\) is doubled: \[ Rate_{new} = k(2[A]) = 2k[A] \] Step 3: Compare with original rate.
\[ \frac{Rate_{new}}{Rate_{old}} = \frac{2k[A]}{k[A]} = 2 \] Conclusion:
For a first-order reaction, doubling the concentration of the reactant doubles the rate of reaction.