We’re being asked to **how the plot will appear if you start with an initial concentration of CO that is much higher than the initial concentration of NO _{2} ([CO]0 >> [NO_{2}]_{0}) then a plot of ln[NO_{2}]**

NO_{2} + CO → NO + CO_{2}

The empirical rate law for this reaction rate = k[NO_{2}]^{2 }

The ** integrated rate law** for a first-order reaction is as follows:

$\overline{){\mathbf{ln}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{t}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{\mathbf{kt}}{\mathbf{}}{\mathbf{+}}{\mathbf{\hspace{0.17em}}}{\mathbf{ln}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{0}}}{\mathbf{}}}$

The ** integrated rate law** for a second-order reaction is as follows:

$\overline{)\frac{\mathbf{1}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{t}}}{\mathbf{=}}{\mathbf{kt}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}}}$

where:

**[A] _{t}** = concentration at time t

**k** = rate constant

**t** = time

**[A] _{0}** = initial concentration

The reaction NO2 + CO → NO + CO_{2} is found to obey the following empirical rate law: rate = k[NO_{2}]^{2 }For this reaction, if you start with an initial concentration of CO that is much higher than the initial concentration of NO_{2} ([CO]0 >> [NO_{2}]_{0}) then a plot of ln[NO_{2}] will appear

A. curved as the kinetics are 2nd order in NO_{2.}

B. linear as the kinetics are 1st order in NO_{2}.

C. linear as the kinetics are pseudo-first order in NO_{2}.

D. curved as the kinetics are 1st order in NO_{2}.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

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