Question

The rate law for the reaction A + B → product is rate = k[A][B]. When will the rate of reaction increase by factor two?

• A. [A] and [B] both are doubled
• B. [A] is doubled and [B] is kept constant
• C. [B] is doubled and [A] is halved
• D. [A] is kept constant [B] is halved

Hint:

In a rate law, if a reactant's concentration is multiplied by a factor 'n' and its order is 'x', the rate changes by n^x. For rate = k[A]^1[B]^1, doubling [A] doubles the rate.

Answer 1

The Correct Option is A

Concept:
The rate law for the reaction is: \[ \text{Rate}=k[A][B] \] where:

• $k$ = rate constant
• $[A]$ = concentration of reactant A
• $[B]$ = concentration of reactant B

This means:

• First order with respect to $A$
• First order with respect to $B$
• Overall order = 2

Step 1: Let original rate be
\[ r = k[A][B] \] We check each option.
Step 2: Option A: Double both $[A]$ and $[B]$
\[ r' = k(2[A])(2[B]) \] \[ r' = 4k[A][B]=4r \] Rate becomes four times.
Step 3: Option B: Double $[A]$, keep $[B]$ constant
\[ r' = k(2[A])[B] \] \[ r' = 2k[A][B]=2r \] Rate becomes twice.
Step 4: Option C: Double $[B]$, half $[A]$
\[ r' = k\left(\frac{[A]}{2}\right)(2[B]) \] \[ r'=k[A][B]=r \] Rate remains same.
Step 5: Option D: Keep $[A]$ constant, half $[B]$
\[ r' = k[A]\left(\frac{[B]}{2}\right) \] \[ r'=\frac{r}{2} \] Rate becomes half.
Step 6: Final Answer
The rate increases by a factor of two when: \[ \boxed{[A]\text{ is doubled and }[B]\text{ is unchanged \] Quick Tip:
In first-order dependence, doubling concentration doubles contribution to rate.