# Chapter 12 - Chemical Kinetics - Exercises: 29

a) $k[NO]^2[Cl_2]$ b) $1.8\times 10^2\frac{L^2}{mol^2\times min}$

#### Work Step by Step

a) Given the data in the table, we can see that doubling the concentration of $Cl_2$ also doubles the initial rate. Solving for n in $2^n=2$ gives a value of $1$ for n, so the $[Cl]$ factor in the rate law is raised to the first power. We can also see that doubling the concentration of $NO$ quadruples the initial rate, Solving for n in $2^n=4$ gives a value of $2$ for n, so the $[NO]$ factor in the rate law is raised to the second power. b) Plugging the first set of results into the rate law found in part (a) gives $0.18\space \frac{mol}{L\times min}=k\times (0.10\space \frac{mol}{L})^2\times 0.10\space \frac{mol}{L}$. Distributing and simplifying the right side of the equation gives $0.18\space \frac{mol}{L\times min}=k\times 0.001\space \frac{mol^3}{L^3}$. Dividing both sides by $0.001\space \frac{mol^3}{L^3}$ yields a value of $1.8\times 10^2\frac{L^2}{mol^2\times min}$. There are 2 significant figures in the answer because all data points given had 2 significant figures.

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