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6) At $500^{\circ} \mathrm{C}$, cyclopropane $\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)$ rearranges to propene $\left(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\right)$. The reaction is 1st order with a rate constant $k=6.7 \times 10^{-4} \mathrm{~s}^{-1}$.
a. Write the rate equation, and determine the rate of reaction when the conversion $x$ is 0 and 0.5 , given that the initial concentration of
$\left[3.35 \times 10-5 \mathrm{~mol} \mathrm{I}^{-1} \mathrm{~s}^{-1} ; 1.675 \times 10^{-5} \mathrm{~mol} \mathrm{I}^{-1} \mathrm{~s}^{-1}\right]$
b. Integrate the rate equation and determine:
i. The concentration of cyclopropane after 30 minutes
$[0.015 \mathrm{M}]$
ii. The time taken for the concentration of cyclopropane to fall to $0.001 \mathrm{~mol}^{2} \mathrm{litr}^{-1}$.
$[5839 \mathrm{~s},(97 \text { minutes)] }$
iii. The fractional conversion of cyclopropane and the concentration of propene after 50 minutes.
[0.866; $0.0433 \mathrm{M}]$
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Answer from Sia

Posted 3 months ago

Solution

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1 Answer

The rate of reaction at conversion $x = 0$ is $3.35 \times 10^{-5} \text{ mol L}^{-1}\text{s}^{-1}$ and at conversion $x = 0.5$ is $1.675 \times 10^{-5} \text{ mol L}^{-1}\text{s}^{-1}$.

2 Answer

The concentration of cyclopropane after 30 minutes is $0.015 \text{ M}$.

3 Answer

The time taken for the concentration of cyclopropane to fall to $0.001 \text{ M}$ is 5839 seconds (97 minutes).

4 Answer

The fractional conversion of cyclopropane after 50 minutes is 0.866, and the concentration of propene is $0.0433 \text{ M}$.

Key Concept

The rate equation for a first-order reaction relates the rate of reaction to the concentration of the reactant and the rate constant.

Explanation

The rate of a first-order reaction decreases as the concentration of the reactant decreases, which is reflected in the rate of reaction at different conversion levels.

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