Determining the Rate Law

The general form of the rate law for the reaction $A + B \rightarrow P$ can be written as:

$r = k[A]^x[B]^y$

where $r$ is the reaction rate, $k$ is the rate constant, $[A]$ and $[B]$ are the concentrations of reactants A and B, and $x$ and $y$ are the reaction orders with respect to A and B, respectively.

Analyzing Concentration Effects on Rate

We use the given information to find the values of $x$ and $y$.

• Condition 1: Effect of doubling $[B]$

  When the concentration of B is doubled ($[B] \rightarrow 2[B]$) while $[A]$ is kept constant, the rate doubles ($r \rightarrow 2r$).

  Initial rate: $r_1 = k[A]^x[B]^y$

  New rate: $r_2 = k[A]^x(2[B])^y$

  Given $r_2 = 2r_1$:

  $k[A]^x(2[B])^y = 2(k[A]^x[B]^y)$

  $2^y = 2$

  Therefore, $y = 1$.

• Condition 2: Effect of doubling both $[A]$ and $[B]$

  When the concentrations of both A and B are doubled ($[A] \rightarrow 2[A]$ and $[B] \rightarrow 2[B]$), the rate increases by a factor of 8 ($r \rightarrow 8r$).

  Initial rate: $r_1 = k[A]^x[B]^y$

  New rate: $r_3 = k(2[A])^x(2[B])^y$

  Given $r_3 = 8r_1$:

  $k(2[A])^x(2[B])^y = 8(k[A]^x[B]^y)$

  $2^x \cdot 2^y = 8$

  Substitute $y = 1$:

  $2^x \cdot 2^1 = 8$

  $2^{x+1} = 2^3$

  $x+1 = 3$

  Therefore, $x = 2$.

Writing the Rate Law

Substituting the determined orders $x=2$ and $y=1$ into the general rate law:

$r = k[A]^2[B]^1$

$r = k[A]^2[B]$

This rate law corresponds to Option B.