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Chapter 18: Problem 88

The degradation of \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of \(k=2.1 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(10^{\circ} \mathrm{C}\). If the tropospheric concentrations of \(\mathrm{OH}\) and \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) are \(1.0 \times 10^{12}\) and \(7.5 \times 10^{14}\) molecules \(/ \mathrm{m}^{3}\), respectively, what is the rate of reaction at this temperature in \(M /\) s?

Short Answer

Expert verified
The rate of reaction at 10°C is approximately \(4.36 \times 10^{-33} M/s\).

Step by step solution

01

Convert the concentration from molecules/m³ to Molarity

First, we need to convert the concentration from molecules/m³ to moles/liter (M). We'll use Avogadro's number, 6.022 x 10²³ molecules/mol. \(OH\: concentration\: (M) = \frac{1.0 \times 10^{12}\: molecules/m^3}{6.022 \times 10^{23}\: molecules/mol × 10^3\: liters/m^3} = 1.66 \times 10^{-13} M\) \(CF_3CH_2F\: concentration\: (M) = \frac{7.5 \times 10^{14}\: molecules/m^3}{6.022 \times 10^{23} \: molecules/mol × 10^3\: liters/m^3} = 1.25 \times 10^{-11} M\)
02

Calculate the rate of reaction using the rate law

We are given that the decay is first order with respect to both reactants. This means that the rate law is equal to the product of the rate constant and the concentration of each reactant. Rate of reaction = k [\(OH\)] \([CF_3CH_2F]\) Now we can plug in the values we have for \(k\), [\(OH\)] and [\(CF_3CH_2F\)] to calculate the rate of reaction: Rate of reaction = \((2.1 \times 10^{8} M^{-1} s^{-1})(1.66 \times 10^{-13} M)(1.25 \times 10^{-11} M)\)
03

Calculate the rate of reaction in M/s

Multiply the given values to find the rate of reaction: Rate of reaction = \(4.36 \times 10^{-33} M/s\) The rate of reaction at 10°C is approximately \(4.36 \times 10^{-33} M/s\).

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