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

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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

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

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(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 $\frac{d[NOCl]}{dt}$?

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