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The rate constant (\mathrm{k})for a certain reaction is 0.05 \mathrm{~s}^1 \text{at} \; 25^{\circ} \mathrm{C}. If the reaction is carried out at 35^{\circ} \mathrm{C} and the effective activation energy (Ea) is \frac{50 \mathrm{~kJ}}{\mathrm{~mol}}, what will be the approximate value of the new rate constant \left(\mathrm{k}^{\prime}\right) at the higher temperature?

Option: 1

\begin{aligned} & 0.015 s^{-1} \\ \end{aligned}


Option: 2

\begin{aligned} & 0.025 s^{-1} \\ \end{aligned}


Option: 3

\begin{aligned} & 0.035 s^{-1} \\ \end{aligned}


Option: 4

\begin{aligned} & 0.045 s^{-1} \end{aligned}


Answers (1)

best_answer

To determine the new rate constant \left(\mathrm{k}^{\prime}\right) at the higher temperature, we can use the
Arrhenius equation:
k^{\prime}=k\ \exp \left[\left(\frac{E a}{R}\right) *\left(\left(\frac{1}{T^{\prime}}\right)-\left(\frac{1}{T}\right)\right)\right]
where \mathrm{k} is the rate constant at the initial temperature \left(25^{\circ} \mathrm{C}\right), \mathrm{Ea} is the effective activation energy, \mathrm{R} is the gas constant \left(\frac{8.314 \mathrm{~J}}{(\mathrm{~mol} \cdot \mathrm{K})}\right), \mathrm{T} is the initial temperature in Kelvin, and \mathrm{T}^{\prime} is the final temperature in Kelvin.

Given:

\begin{aligned} & \mathrm{k}=0.05 \mathrm{~s}^{-1}, \\ \\& \mathrm{Ea}=\frac{50 \mathrm{~kJ}}{\mathrm{~mol}}, \\ \\& T=25^{\circ} \mathrm{C}=298 \mathrm{~K}, \\ \\& T^{\prime}=35^{\circ} \mathrm{C}=308 \mathrm{~K} . \end{aligned}

Plugging in the values into the equation, we have:

\mathrm{k^{\prime}=0.05 \times \exp \left[\left(\frac{50000 J}{m o l} /\left(\frac{8.314 J}{(m o l \cdot K)}\right)\right) \times \left(\left(\frac{1}{308 K}\right)-\left(\frac{1}{298 K}\right)\right)\right]}

Calculating the right-hand side of the equation gives:

k^{\prime} \approx 0.025 \mathrm{~s}^{-1}

Therefore, the approximate value of the new rate constant \left(k^{\prime}\right) at the higher temperature is 0.025 \mathrm{~s}^{-1}, which corresponds to option 2.

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