7) The compound $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$ decomposes to $\mathrm{C}_{2} \mathrm{H}_{4}$ and $\mathrm{HCl}$ at elevated temperatures. Find the rate law for this reaction from the following rate data.
\begin{tabular}{ll}
P$_{\mathrm{CH} 3 \mathrm{CH} 2 \mathrm{Cl}}$ & Initial rate (atm/sec) \
$1.8 imes 10^{-2}$ & $7.5 imes 10^{-4}$ \
$4.3 imes 10^{-2}$ & $1.8 imes 10^{-3}$ \
$7.8 imes 10^{-2}$ & $3.3 imes 10^{-3}$
\end{tabular}

Question

7) The compound $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$ decomposes to $\mathrm{C}_{2} \mathrm{H}_{4}$ and $\mathrm{HCl}$ at elevated temperatures. Find the rate law for this reaction from the following rate data.
\begin{tabular}{ll} 
P$_{\mathrm{CH} 3 \mathrm{CH} 2 \mathrm{Cl}}$ & Initial rate (atm/sec) \
$1.8 	imes 10^{-2}$ & $7.5 	imes 10^{-4}$ \
$4.3 	imes 10^{-2}$ & $1.8 	imes 10^{-3}$ \
$7.8 	imes 10^{-2}$ & $3.3 	imes 10^{-3}$
\end{tabular}

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the reaction**: The given reaction is the decomposition of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$ to $\mathrm{C}_{2} \mathrm{H}_{4}$ and $\mathrm{HCl}$. ‖ ‖ 2 ⋮ **Write the general rate law**: The rate law for the reaction can be written as $\text{Rate} = k [\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}]^n$, where $k$ is the rate constant and $n$ is the order of the reaction with respect to $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$. ‖ ‖ 3 ⋮ **Use the given data to determine the order of the reaction**: - For $P_{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}} = 1.8 \times 10^{-2}$ atm, the initial rate is $7.5 \times 10^{-4}$ atm/sec. - For $P_{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}} = 4.3 \times 10^{-2}$ atm, the initial rate is $1.8 \times 10^{-3}$ atm/sec. - For $P_{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}} = 7.8 \times 10^{-2}$ atm, the initial rate is $3.3 \times 10^{-3}$ atm/sec. To find $n$, we can use the ratio of the rates and the concentrations: $$\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}]_2}{[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}]_1}\right)^n$$ Using the first two data points: $$\frac{1.8 \times 10^{-3}}{7.5 \times 10^{-4}} = \left(\frac{4.3 \times 10^{-2}}{1.8 \times 10^{-2}}\right)^n$$ $$2.4 = \left(2.39\right)^n$$ Taking the logarithm of both sides: $$\log(2.4) = n \log(2.39)$$ $$n \approx 1$$ Thus, the reaction is first order with respect to $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$. ‖ ‖ 4 ⋮ **Write the rate law**: The rate law for the reaction is $\text{Rate} = k [\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}]$. ‖ ∻Answer∻ ⚹ The rate law for the decomposition of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$ is $\text{Rate} = k [\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}]$. ⚹ ∻Key Concept∻ ⚹ Rate law determination ⚹ ∻Explanation∻ ⚹ The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. By comparing the initial rates and concentrations, we determined that the reaction is first order with respect to $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$. ⚹◊What is the order of the reaction with respect to $CH_{3}CH_{2}Cl$ in this given data?⍭ Generate me a similar question◊

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