Recall that the **rate law** only focuses on the reactant concentrations and has a general form of:

$\overline{){\mathbf{rate}}{\mathbf{}}{\mathbf{law}}{\mathbf{=}}{\mathbf{k}}{\left[\mathbf{A}\right]}^{{\mathbf{x}}}{\left[\mathbf{B}\right]}^{{\mathbf{y}}}}$

k = rate constant

A & B = reactants

x & y = reactant orders

**Step 1.**** Calculate the order of the reaction with respect to N _{2}O_{5}.**

Use experimental data 1 and 2 (you can use any):

$\frac{\mathbf{rate}\mathbf{}\mathbf{2}}{\mathbf{rate}\mathbf{}\mathbf{1}}\mathbf{=}\frac{{\displaystyle {\left[{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\right]}^{\mathbf{x}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{rate}\mathbf{}\mathbf{2}}}{{\displaystyle {\left[{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\right]}^{\mathbf{x}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{rate}\mathbf{}\mathbf{1}}}$

*x = order*

*(Larger concentration should be on the numerator)*

**Solve for x:**

$\frac{\mathbf{2}\mathbf{.}\mathbf{26}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{8}\mathbf{.}\mathbf{90}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}\mathbf{=}\frac{{\left[\mathbf{0}\mathbf{.}\mathbf{190}\right]}^{\mathbf{x}}}{{\left[\mathbf{0}\mathbf{.}\mathbf{0750}\right]}^{\mathbf{x}}}$

*since the numerator and the denominator are raised to the same power, you can simplify the equation to:*

$\frac{\mathbf{2}\mathbf{.}\mathbf{26}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{8}\mathbf{.}\mathbf{90}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}\mathbf{=}{\left[\frac{\mathbf{0}\mathbf{.}\mathbf{190}}{\mathbf{0}\mathbf{.}\mathbf{0750}}\right]}^{\mathbf{x}}\phantom{\rule{0ex}{0ex}}\mathbf{2}\mathbf{.}\mathbf{53}\mathbf{=}{\left[\mathbf{2}\mathbf{.}\mathbf{53}\right]}^{\mathbf{x}}$

*2.53 raised to the power of 1 = 2.53*

*x = 1 → 1st **order with respect to N_{2}O_{5}*

**Step 2****. Determine the rate law of the reaction**

$\overline{){\mathbf{rate}}{\mathbf{}}{\mathbf{law}}{\mathbf{=}}{\mathbf{k}}{\left[\mathbf{A}\right]}^{{\mathbf{x}}}{\left[\mathbf{B}\right]}^{{\mathbf{y}}}}$

*Substitute N_{2}O_{5} and O_{2} and their orders:*

**rate law = k[ N_{2}O_{5}]^{1}**

*or simply*

**rate = k[ N_{2}O_{5}]**

**Step 3. Calculate the value of the rate constant**

The following data were obtained for the gas‑phase decomposition of dinitrogen pentoxide,

2 N_{2}O_{5 }(g) → 4 NO_{2} (g) + O_{2 }(g)

Defining the rate as -Δ[N_{2}O_{5}] / Δt, write the rate law and calculate the value of the rate constant.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.