$$\text{Activation energy} = 250\,\text{kJ} - 50\,\text{kJ} = 200\,\text{kJ}$$

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

Question 1

Based on the provided reaction pathway diagram, what is the activation energy for the forward reaction?

Accepted Answer

$$	ext{Activation energy} = 250\,	ext{kJ} - 50\,	ext{kJ} = 200\,	ext{kJ}$$

Question 2

What is the enthalpy change ($$\Delta H) $$for the reaction shown in the diagram?

Accepted Answer

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

Question 3

Would the $$\Delta H$$ value for the reverse reaction be the same magnitude as for the forward reaction, and what would its sign be?

Accepted Answer

Yes, the magnitude would be the same, but the sign would be negative: $$\Delta H = -50\,	ext{kJ}$$

Question 4

If a catalyst is added to this reaction, which of the following changes would occur in the diagram?

Accepted Answer

The peak of the curve (activation energy) would decrease, but $$\Delta H$$ would remain unchanged.

Analyzing Activation Energy and Enthalpy from a Reaction Energy Diagram

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