Question

Suppose that a catalyst lowers the activation barrier of a reaction from 126 kJ/mol to 52 kJ/mol . Part A By what factor would you expect the reaction rate to increase at 25 ∘C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)

Answer 1

Answer 2

To calculate the factor by which the reaction rate increases due to the catalyst, we use the Arrhenius equation:

$$k=A{e}^{-\frac{E_a}{RT}}$$

Where:

• $k$ = rate constant

• $A$ = frequency factor (same for both reactions)

• $E_a$ = activation energy

• $R$ = gas constant ()

• $T$ = temperature in Kelvin ($25\mathrm{°}C=298\text{K}$)

• 
  

Step 1: Calculate the ratio of rate constants ($\frac{k_{\text{catalyzed}}}{k_{\text{uncatalyzed}}}$$\frac{k_{\text{catalyzed}}}{k_{\text{uncatalyzed}}}$)

$$\frac{k_{\text{cat}}}{k_{\text{uncat}}}=\frac{e^{-\frac{E_{a,\text{cat}}}{RT}}}{e^{-\frac{E_{a,\text{uncat}}}{RT}}}=e^{\frac{E_{a,\text{uncat}}-E_{a,\text{cat}}}{RT}}$$

Step 2: Plug in the given values

$$E_{a,\text{uncat}}=126\text{kJ/mol}=126,000\text{J/mol}$$$$E_{a,\text{cat}}=52\text{kJ/mol}=52,000\text{J/mol}$$$$\mathrm{\Delta }{E}_a=126,000-52,000=74,000\text{J/mol}$$

Now, compute the exponent:

$$\frac{\mathrm{\Delta }{E}_a}{RT}=\frac{74,000}{8.314\times 298}\approx 29.83$$

Step 3: Calculate the factor

$$\frac{k_{\text{cat}}}{k_{\text{uncat}}}=e^{29.83}\approx 8.9\times 10^{12}$$

Final Answer:

The reaction rate increases by a factor of $8.9\times 10^{12}$ at $25\mathrm{°}C$