## Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning

# Chapter 11 - Exercises - Page 487b: 29

#### Answer

a) $Rate= k[NO]^{2}[Cl_{2}]$ b) $180\, L^{2}/mol^{2}\cdot min$

#### Work Step by Step

a) $Rate= k[NO]^{x}[Cl_{2}]^{y}$ When the concentration of NO is doubled and that of $Cl_{2}$ is kept constant, then the initial rate increases by a factor of four from 0.36 to 1.45 This implies that the rate depends upon the square of [NO]. When the concentration of $Cl_{2}$ is doubled and that of NO is kept constant, the rate also gets doubled. This implies that the rate depends on the first power of $[Cl_{2}]$. Hence, $Rate= k[NO]^{2}[Cl_{2}]$ b) $k= \frac{Rate}{[NO]^{2}[Cl_{2}]}=\frac{0.18\,mol/L\cdot min}{(0.10\,mol/L)^{2}\times0.10\,mol/L}=180\, L^{2}/mol^{2}\cdot min$

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