Question
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?
Step 1
The equation is given by: \[k = A \exp\left(-\frac{Ea}{RT}\right)\] where A is the pre-exponential factor, R is the gas constant, and T is the temperature in Kelvin. Show more…
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A chemical reaction has an activation energy of 30. 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.
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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Topic 2 : Effect of Temperature and Catalyst on Rate of Reactions, Collosion Theory of Chemical Reactions
The rates of many reactions approximately double for each $10^{\circ} \mathrm{C}$ rise in temperature. Assuming a starting temperature of $25^{\circ} \mathrm{C}$, what would the activation energy be, in $\mathrm{kJ} \mathrm{mol}^{-1}$, if the rate of a reaction were to be twice as large at $35^{\circ} \mathrm{C} ?$
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