## Related questions with answers

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

Solution

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

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

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

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 $20~^\circ\text{C}$:

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

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

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