$$E_a = 250\ \text{kJ} - 50\ \text{kJ} = 200\ \text{kJ}$$

$$\Delta H = 100\ \text{kJ} - 50\ \text{kJ} = +50\ \text{kJ}$$

$$E_a = 250\ \text{kJ} - 100\ \text{kJ} = 150\ \text{kJ}$$

Question 1

Refer to the reaction energy diagram provided. What is the activation energy ($$E_a) $$for the forward reaction?

Accepted Answer

$$E_a = 250\ 	ext{kJ} - 50\ 	ext{kJ} = 200\ 	ext{kJ}$$

Question 2

Based on the reaction energy diagram, what is the enthalpy change ($$\Delta H) $$for the reaction?

Accepted Answer

$$\Delta H = 100\ 	ext{kJ} - 50\ 	ext{kJ} = +50\ 	ext{kJ}$$

Question 3

If the reaction shown in the diagram is reversed, what would be the sign and value of $$\Delta H$$ for the reverse reaction?

Accepted Answer

$$\Delta H = -50\ 	ext{kJ}$$

Question 4

What is the activation energy for the reverse reaction based on the diagram?

Accepted Answer

$$E_a = 250\ 	ext{kJ} - 100\ 	ext{kJ} = 150\ 	ext{kJ}$$

Analyzing Activation Energy and Enthalpy from a Reaction Energy Diagram

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