Question

The rate for the following reaction is given by : \[ A+B \rightarrow C \] \[ \text{Rate}=k[A][B]^2 \] (i) How is the rate of reaction affected if we double the concentration of \(B\) ? (ii) Write the overall order of a reaction if \(A\) is present in large excess.

Hint:

For a rate law \[ \text{Rate}=k[A]^m[B]^n \] if the concentration of \(B\) is doubled, the rate changes by a factor of \[ 2^n \] Here, \[ n=2 \] so \[ 2^2=4 \] and the rate becomes four times. If a reactant is present in large excess, its concentration is treated as constant while determining the effective order.

Answer 1

Concept: The rate law of a reaction expresses the dependence of reaction rate on the concentration of reactants. For the given reaction, \[ A+B \rightarrow C \] the experimentally determined rate equation is \[ \text{Rate}=k[A][B]^2 \] where \[ k=\text{rate constant} \] \[ [A]=\text{concentration of reactant }A \] \[ [B]=\text{concentration of reactant }B \] The exponent of each concentration term represents the order with respect to that reactant. Thus, \[ \text{Order with respect to }A=1 \] and \[ \text{Order with respect to }B=2 \]

Part (i)

Step 1: Writing the original rate expression. Initially, \[ R_1=k[A][B]^2 \] where \(R_1\) is the initial rate of reaction.

Step 2: Doubling the concentration of \(B\). When the concentration of \(B\) is doubled, \[ [B] \rightarrow 2[B] \] Substituting this new concentration into the rate law, \[ R_2=k[A](2[B])^2 \] \[ R_2=k[A]\times4[B]^2 \] \[ R_2=4k[A][B]^2 \] Since \[ R_1=k[A][B]^2 \] therefore, \[ R_2=4R_1 \]

Step 3: Interpreting the result. The new rate is four times the original rate. Hence, doubling the concentration of \(B\) increases the rate by a factor of four. \[ \boxed{\text{Rate becomes four times}} \]

Part (ii)

Step 4: Considering \(A\) in large excess. When \(A\) is present in very large excess, its concentration changes negligibly during the reaction. Therefore, \([A]\) can be treated as approximately constant. The rate equation becomes \[ \text{Rate}=k[A][B]^2 \] Let \[ k'=k[A] \] Since \(k\) and \([A]\) are constants under these conditions, \[ k'=\text{constant} \] Hence, \[ \text{Rate}=k'[B]^2 \]

Step 5: Determining the effective order. The modified rate equation contains only \[ [B]^2 \] Therefore, the reaction behaves as a second-order reaction. \[ \boxed{\text{Overall order}=2} \] This situation is commonly referred to as a pseudo-second-order reaction because one reactant is present in large excess and its concentration remains effectively constant.

Final Answers: \[ \boxed{\text{(i) Rate becomes four times}} \] \[ \boxed{\text{(ii) Overall order}=2} \]