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The rate constant at 325C325^{\circ} \mathrm{C} for the decomposition reaction C4H82C2H4\mathrm{C}_4 \mathrm{H}_8 \longrightarrow 2 \mathrm{C}_2 \mathrm{H}_4 is 6.1×108 s6.1 \times 10^{-8} \mathrm{~s}, and the activation energy is 261 kJ261 \mathrm{~kJ} per mole of C4H8\mathrm{C}_4 \mathrm{H}_8. Determine the frequency factor for the reaction.

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Answered 2 years ago
Answered 2 years ago
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Assuming that rate of reaction doubles for every 10 ^\circC increase in temperature, the reaction rate will be doubled two times since the temperature difference is 20 C20~^\circ\text{C}. This can be mathemically expressed as:

reaction rate=22\begin{aligned}\text{reaction rate} =2^{2}\end{aligned}

reaction rate=4\begin{aligned}\text{reaction rate} =4\end{aligned}

Therefore, the reaction rate will be 4 times faster. Since the original reaction took 48 mins, we can calculate for the time it will took when the temperature increase is equal to 20 C20~^\circ\text{C}:

time=48 mins4\begin{aligned}\text{time} =\frac{48\text{ mins}}{4}\end{aligned}

time=12 mins\begin{aligned}\text{time} =12\text{ mins}\end{aligned}

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