Question
A chemical reaction has an activation energy of30. kJ/mol. If you had to slow down this reaction a thousandfold by cooling it from room temperature $\left(25^{\circ} \mathrm{C}\right),$ calculate what the temperature would be.
Step 1
The equation is given by: \[k = A \exp\left(-\frac{E_a}{RT}\right)\] where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature. Show more…
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A chemical reaction has an activation energy of $30 . \mathrm{kJ} / \mathrm{mol}$. If you had to slow down this reaction a thousandfold by cooling it from room temperature $\left(25^{\circ} \mathrm{C}\right)$, what would the temperature be?
The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. The activation energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of the reaction is $.$ Take; $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ In $\overline{3.555}=1.268$
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When the temperature of a reaction increases from $27^{\circ} \mathrm{C}$ to $37^{\circ} \mathrm{C}$, the rate increases by $2.5$ times, the activation energy in the temperature range is: (a) $70.8 \mathrm{~kJ}$ (b) $7.08 \mathrm{~kJ}$ (c) $35.8 \mathrm{~kJ}$ (d) $14.85 \mathrm{~kJ}$
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