If an increase in temperature from 25°C to 35°C doubles the reaction rate constant, what is the activation energy of the reaction? Assume the Arrhenius equation applies.

Chemists commonly use a rule of thumb that an increase in temperature of 10 K doubles the reaction rate. What must the activation energy be for this statement to be true if the change in temperature is from 25°C to 35°C?

Chemists commonly use a rule of thumb that an increase in temperature of 10 K doubles the reaction rate. What must the activation energy be for this statement to be true if the change in temperature is from 25°C to 35°C?

For two measured rate constants for the same reaction at different temperatures: Temperature / °C | k / s⁻¹ - 199 | 4.50 × 10⁻⁵ - 258 | 3.2 × 10⁻³ What is the activation energy (in kJ mol⁻¹) for this reaction?

Using the Arrhenius Equation, if the activation energy for a reaction is 80 kJ/mol, at what temperature will the rate be ten times faster than at 0 degrees Celsius?

Suppose that a catalyst lowers the activation barrier of a reaction from 122 kJ/mol to 54 kJ/mol. By what factor would you expect the reaction rate to increase at 25.00 °C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are the same.)

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower does the activation barrier have to be when sucrose is in the presence of the enzyme?

The activation energy Ea for a particular reaction is 50.0 kJ/mol. How much faster is the reaction at 313 K than at 310.0 K? (R = 8.314 J/mol • K)

The activation energy for a first-order reaction is 26.5 kJ/mol. At 10.0°C, the rate constant is 0.020 s⁻¹. Calculate the temperature at which the rate constant is 0.040 s⁻¹ using the Arrhenius Equation.