**Question 16 of 20****Problem Statement:**What is ΔH (in kJ/mol) for a reaction where the activation energy is 250 kJ/mol, while the activation energy for the reverse reaction is 110.0 kJ/mol?**Explanation:**To find ΔH, use the formula:\[ \Delta H = \text{Activation Energy (forward)} - \text{Activation Energy (reverse)} \]Given:- Activation Energy (forward) = 250 kJ/mol- Activation Energy (reverse) = 110.0 kJ/mol**Calculation:**\[ \Delta H = 250 \, \text{kJ/mol} - 110.0 \, \text{kJ/mol} \]\[ \Delta H = 140.0 \, \text{kJ/mol} \]### Interface Description:- The image shows a calculator interface for inputting the answer in kJ/mol.- Number pad includes digits 0-9, a decimal (.), a +/- button, a backspace (×) button, a clear (C) button, and a multiplication by 10 (x10^□) function.**Note:**For further assistance or resources, tap or pull up the additional options available at the bottom of the interface.

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What is AH (in kJ/mol) for a reaction where the activation energy is 250 kJ/mol while the activation energy for the reverse reaction is 110.0 kJ/mol?

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

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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