Question attached ### Problem StatementA certain reaction has an activation energy of 30.07 kJ/mol. At what Kelvin temperature will the reaction proceed 7.50 times faster than it did at 353 K?**Equation Placeholder:**\[ T = \_\_\_\_\_\_ \, \text{K} \]### ExplanationIn this problem, we are asked to find the temperature at which a reaction proceeds 7.50 times faster than it does at 353 K, given the activation energy of 30.07 kJ/mol. This requires understanding the relationship between the rate of a reaction, temperature, and activation energy, often modeled by the Arrhenius equation:\[ k = A \cdot e^{-\frac{E_a}{RT}} \]Where:- \( k \) is the rate constant.- \( A \) is the pre-exponential factor (frequency of collisions).- \( E_a \) is the activation energy.- \( R \) is the universal gas constant (8.314 J/mol·K).- \( T \) is the temperature in Kelvin.### ObjectiveDetermine the temperature \( T \) such that the rate constant \( k \) at this temperature is 7.50 times greater than the rate constant at 353 K. The calculations involve using the modified form of the Arrhenius equation to solve for \( T \).### Additional Steps1. **Understand the Relationship**: The rate of reaction depends exponentially on temperature. Increasing the temperature generally increases the rate.2. **Perform Calculations**: Use the relationship \(\frac{k_2}{k_1} = 7.50\) to solve for the new temperature \( T \).### ConclusionOnce the temperature is calculated using the Arrhenius equation and appropriate transformations for the given rate increase, it can be plugged into the equation placeholder.

Answered: A certain reaction has an activation…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Question attached

### Problem Statement

A certain reaction has an activation energy of 30.07 kJ/mol. At what Kelvin temperature will the reaction proceed 7.50 times faster than it did at 353 K?

**Equation Placeholder:**

\[ T = \_\_\_\_\_\_ \, \text{K} \]

### Explanation

In this problem, we are asked to find the temperature at which a reaction proceeds 7.50 times faster than it does at 353 K, given the activation energy of 30.07 kJ/mol. This requires understanding the relationship between the rate of a reaction, temperature, and activation energy, often modeled by the Arrhenius equation:

\[ k = A \cdot e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) is the rate constant.
- \( A \) is the pre-exponential factor (frequency of collisions).
- \( E_a \) is the activation energy.
- \( R \) is the universal gas constant (8.314 J/mol·K).
- \( T \) is the temperature in Kelvin.

### Objective

Determine the temperature \( T \) such that the rate constant \( k \) at this temperature is 7.50 times greater than the rate constant at 353 K. The calculations involve using the modified form of the Arrhenius equation to solve for \( T \).

### Additional Steps

1. **Understand the Relationship**: The rate of reaction depends exponentially on temperature. Increasing the temperature generally increases the rate.
2. **Perform Calculations**: Use the relationship \(\frac{k_2}{k_1} = 7.50\) to solve for the new temperature \( T \).

### Conclusion

Once the temperature is calculated using the Arrhenius equation and appropriate transformations for the given rate increase, it can be plugged into the equation placeholder.

Expert Solution

Step 1

Arrhenius equation:

If we know the value of the rate constant at two different temperatures, and the activation energy we can calculate the temperature of the reaction by using the Arrhenius equation. The Arrhenius equation is given by:

where k₂= rate constant at the T₂ temperature

k₁ = rate constant at T₁

k₂ = 7.50 k₁

E_a = activation energy = 30.07 kJ/mol = 30.07x10³ J/mol

R = gas constant = 8.314 J/K.mol

T₁ = 353 K

T₂ = to be calculate

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