For the reaction 2 NO + Cl_2 rightarrow 2 N | vedclass

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

Question

(B) For the reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$,the rate of reaction is given by: Rate $= -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[Cl_2]}{dt

Vedclass · Accepted Answer

$\frac{d[NO]}{dt} = -\frac{d[NOCl]}{dt}$

Vedclass · Answer

(B) For the reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$,the rate of reaction is given by: Rate $= -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[Cl_2]}{dt} = \frac{1}{2} \frac{d[NOCl]}{dt}$. Since $NO$ is a reactant,its rate of disappearance is $-\frac{d[NO]}{dt}$. Since $NOCl$ is a product,its rate of appearance is $\frac{d[NOCl]}{dt}$. From the rate expression: $-\frac{1}{2} \frac{d[NO]}{dt} = \frac{1}{2} \frac{d[NOCl]}{dt}$. Therefore,$-\frac{d[NO]}{dt} = \frac{d[NOCl]}{dt}$,or $\frac{d[NO]}{dt} = -\frac{d[NOCl]}{dt}$.

For the reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$,what is the relation between $\frac{d[NO]}{dt}$ and $\frac{d[NOCl]}{dt}$?

Similar Questions

The rate of reaction is expressed as
$+\frac{1}{2}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[D]}{dt} = +\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt}$
The reaction is:

For the reaction,$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$,the rate of formation of $O_2$ is $0.032 \ g \ h^{-1}$. What is the rate of conversion of $N_2O_5$ in $g \ h^{-1}$?

Given above is the concentration vs time plot for a dissociation reaction: $A \rightarrow nB$. Based on the data of the initial phase of the reaction (initial $10 \ min$), the value of $n$ is . . . . . . .

For the reaction $2N_2O_{5(g)} \rightleftharpoons 4NO_{2(g)} + O_{2(g)}$ in a closed vessel,the concentration of $NO_2$ increases by $2.0 \times 10^{-2} \, mol \, L^{-1}$ in $5 \, s$. Calculate the rate of reaction.

Identify the correct expression for the rate of the following reaction: $2 N_2 O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module