The rate equation for the reaction 2 A + B lo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The rate law for the reaction is given by $	ext{Rate} = k[A][B]^2$. Given values are: $k = 5.0 	imes 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$ $[A] = 0

Vedclass · Accepted Answer

$2.50 	imes 10^{-9} \ mol \ L^{-1} \ s^{-1}$

Vedclass · Answer

(C) The rate law for the reaction is given by $	ext{Rate} = k[A][B]^2$. Given values are: $k = 5.0 	imes 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$ $[A] = 0.05 \ mol \ L^{-1} = 5 	imes 10^{-2} \ mol \ L^{-1}$ $[B] = 0.1 \ mol \ L^{-1} = 1 	imes 10^{-1} \ mol \ L^{-1}$ Substituting these values into the rate equation: $	ext{Rate} = (5.0 	imes 10^{-6}) 	imes (0.05) 	imes (0.1)^2$ $	ext{Rate} = 5.0 	imes 10^{-6} 	imes 5 	imes 10^{-2} 	imes 1 	imes 10^{-2}$ $	ext{Rate} = 25 	imes 10^{-10} \ mol \ L^{-1} \ s^{-1}$ $	ext{Rate} = 2.50 	imes 10^{-9} \ mol \ L^{-1} \ s^{-1}$

The rate equation for the reaction $2 A + B \longrightarrow$ products is $\text{rate} = k[A][B]^2$. If $k$ at $T \ K$ is $5.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$,the initial rate of the reaction,when $[A] = 0.05 \ mol \ L^{-1}$ and $[B] = 0.1 \ mol \ L^{-1}$ is:

Similar Questions

The order of a reaction for an esterification process is . . . . . . .

Give the relation between half-life $(t_{1/2})$ and initial concentration of reactant $([R]_0)$ for an $(n-1)^{th}$ order reaction.

If the rate of reaction is given by $Rate = K [A]^{3/2} [B]^{-1/2}$,then find the order of the reaction.

The rate of the reaction: $2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$ ; $\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

Which of the following statements regarding the order of a reaction is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module