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.)

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?

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.

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.

The activation energy for the decomposition of HI(g) to H2(g) and I2(g) is 186 kJ/mol. The rate constant at 555 K is 3.52 × 10^-7 L/mol-s. What is the rate constant at 645 K?