The following questions refer to the reaction 2A₂ + B₂ → 2C. The mechanism below has been proposed:
-The activation energy for the reaction H₂(g) + I₂(g) → 2HI(g) is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a Pt catalyst. Calculate the value of the ratio rate(catalyzed) /rate(uncatalyzed) .

Question

The following questions refer to the reaction 2A₂ + B₂ → 2C. The mechanism below has been proposed:

The following questions refer to the reaction 2A<sub>2</sub> + B<sub>2</sub> → 2C. The mechanism below has been proposed: -The activation energy for the reaction H<sub>2</sub>(g) + I<sub>2</sub>(g) → 2HI(g) is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a Pt catalyst. Calculate the value of the ratio rate(catalyzed) /rate(uncatalyzed) . A) 1.00 B) 0.321 C) 1.38 D) 7.62 × 10<sup>10</sup> E) none of these

-The activation energy for the reaction H₂(g) + I₂(g) → 2HI(g) is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a Pt catalyst. Calculate the value of the ratio rate(catalyzed) /rate(uncatalyzed) .

Quizplus · Accepted Answer

The ratio of the rates can be calculated using the Arrhenius equation, which relates the rate constant $k$ to the activation energy $E_a$, the gas constant $R$, and the temperature $T$. The equation is given by $k = A e^{-E_a/(RT)}$, where $A$ is the pre-exponential factor (assumed to be constant for both the catalyzed and uncatalyzed reactions). The ratio of the rates for the catalyzed and uncatalyzed reactions can be found by taking the ratio of their respective rate constants, which, due to the exponential nature of the equation, depends only on the change in activation energy and not on $A$.Given:- $E_{a1} = 184 \, \text{kJ/mol}$ (uncatalyzed)- $E_{a2} = 59.0 \, \text{kJ/mol}$ (catalyzed)- $R = 8.314 \times 10^{-3} \, \text{kJ/mol}\cdot K$ (gas constant)- $T = 600 \, K$The ratio of the rates is given by:$$\frac{\text{rate(catalyzed)}}{\text{rate(uncatalyzed)}} = \frac{e^{-E_{a2}/(RT)}}{e^{-E_{a1}/(RT)}}$$Substituting the given values:$$\frac{e^{-59.0/(8.314 \times 10^{-3} \times 600)}}{e^{-184/(8.314 \times 10^{-3} \times 600)}}$$Calculating the exponentials:$$\frac{e^{-59.0/(4.9884)}}{e^{-184/(4.9884)}}$$This results in a ratio of rates significantly greater than 1, indicating that the catalyzed reaction is much faster than the uncatalyzed one. The exact calculation of this expression gives a value of approximately $7.62 \times 10^{10}$, which matches choice D.

The Following Questions Refer to the Reaction 2A2 + B2

The Following Questions Refer to the Reaction 2A₂ + B₂