In a reaction, $A + B \to C$ , the rate expression is $R = k [A] [B]^{2}$ . If the concentration of both the reactant is doubled at constant volume, then the rate of the reaction will be:

eight times

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Solution

The correct option is B eight times
Given, $R = k [A] [B]^{2}$
when $[A]^{'} = [2 A]$ and $[B]^{'} = [2 B]$
$R^{'} = k [2 A] [2 B]^{2}$
$= k [2 A] 4 [B]^{2} = 8 k [A] [B]^{2}$
$\Rightarrow R^{'} = 8 R$
Thus, the rate will become eight times.

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In a reaction, A+B→C, the rate expression is R=k[A][B]2. If the concentration of both the reactant is doubled at constant volume, then the rate of the reaction will be:

The correct option is B eight timesGiven, R=k[A][B]2when [A]′=[2A] and [B]′=[2B]R′=k[2A][2B]2=k[2A]4[B]2=8k[A][B]2⇒R′=8RThus, the rate will become eight times.

In a reaction, $A + B \to C$ , the rate expression is $R = k [A] [B]^{2}$ . If the concentration of both the reactant is doubled at constant volume, then the rate of the reaction will be:

The correct option is B eight times
Given, $R = k [A] [B]^{2}$
when $[A]^{'} = [2 A]$ and $[B]^{'} = [2 B]$
$R^{'} = k [2 A] [2 B]^{2}$
$= k [2 A] 4 [B]^{2} = 8 k [A] [B]^{2}$
$\Rightarrow R^{'} = 8 R$
Thus, the rate will become eight times.