For a reaction $2A + B \to $ Products, doubling the initial concentration of both the reactants increases the rate by a factor of $8$, and doubling the concentration of $+B$ alone doubles the rate. The rate law for the reaction is
For a reaction $2A + B \to $ Products, doubling the initial concentration of both the reactants increases the rate by a factor of $8$, and doubling the concentration of $+B$ alone doubles the rate. The rate law for the reaction is
- A
$\gamma = k[A]{[B]^2}$
- B
$\gamma = k{[A]^2}[B]$
- C
$\gamma = k[A][B]$
- D
$\gamma = k{[A]^2}{[B]^2}$
Similar Questions
The rate law of the reaction $A + 2B \to $Product is given by $\frac{{d[dB]}}{{dt}} = k[{B^2}]$. If $ A$ is taken in excess, the order of the reaction will be
The rate law of the reaction $A + 2B \to $Product is given by $\frac{{d[dB]}}{{dt}} = k[{B^2}]$. If $ A$ is taken in excess, the order of the reaction will be
$A_2 + 2\,B \to 2\,AB$
$[A_2]$
$[B]$
${-d\,[A_2]/dt}$
$0.1$
$0.2$
$1 \times {10^{ - 2}}\,M{s^{ - 1}}$
$0.2$
$0.2$
$2 \times {10^{ - 2}}\,M{s^{ - 1}}$
$0.2$
$0.4$
$8 \times {10^{ - 2}}\,M{s^{ - 1}}$
Order of reaction w.r.t. $A_2$ and $B$ are respectively
$A_2 + 2\,B \to 2\,AB$
| $[A_2]$ | $[B]$ | ${-d\,[A_2]/dt}$ |
| $0.1$ | $0.2$ | $1 \times {10^{ - 2}}\,M{s^{ - 1}}$ |
| $0.2$ | $0.2$ | $2 \times {10^{ - 2}}\,M{s^{ - 1}}$ |
| $0.2$ | $0.4$ | $8 \times {10^{ - 2}}\,M{s^{ - 1}}$ |
Order of reaction w.r.t. $A_2$ and $B$ are respectively
Which one of the following statements is wrong
Which one of the following statements is wrong
The hypothetical reaction : $2A + B \to C + D$ is catalyzed by $E$ as indicated in the possible mechanism below -
Step$-1$ : ${\text{A + E }} \rightleftharpoons AE$ (fast)
Step$-2$ :${\text{AE + A }} \to {A_2} + E$ (slow)
Step$-3$ :${{\text{A}}_2}{\text{ + B }} \to {\text{D}}$ (fast)
what rate law best agrees with this mechanism
The hypothetical reaction : $2A + B \to C + D$ is catalyzed by $E$ as indicated in the possible mechanism below -
Step$-1$ : ${\text{A + E }} \rightleftharpoons AE$ (fast)
Step$-2$ :${\text{AE + A }} \to {A_2} + E$ (slow)
Step$-3$ :${{\text{A}}_2}{\text{ + B }} \to {\text{D}}$ (fast)
what rate law best agrees with this mechanism
For the reaction $A + B \to $ products, what will be the order of reaction with respect to $A$ and $B$ ?
Exp.
$[A]\,(mol\,L^{-1})$
$[B]\,(mol\,L^{-1})$
Initial rate $(mol\,L^{-1}\,s^{-1})$
$1.$
$2.5\times 10^{-4}$
$3\times 10^{-5}$
$5\times 10^{-4}$
$2.$
$5\times 10^{-4}$
$6\times 10^{-5}$
$4\times 10^{-3}$
$3.$
$1\times 10^{-3}$
$6\times 10^{-5}$
$1.6\times 10^{-2}$
For the reaction $A + B \to $ products, what will be the order of reaction with respect to $A$ and $B$ ?
| Exp. | $[A]\,(mol\,L^{-1})$ | $[B]\,(mol\,L^{-1})$ | Initial rate $(mol\,L^{-1}\,s^{-1})$ |
| $1.$ | $2.5\times 10^{-4}$ | $3\times 10^{-5}$ | $5\times 10^{-4}$ |
| $2.$ | $5\times 10^{-4}$ | $6\times 10^{-5}$ | $4\times 10^{-3}$ |
| $3.$ | $1\times 10^{-3}$ | $6\times 10^{-5}$ | $1.6\times 10^{-2}$ |