Question

1. consider the reaction c + d → cd.

the (ea_(fwd)) is +65.0 kj and the ea_(rev) is +150 kj. draw and label a potential energy diagram for this reaction. calculate and label δh°_r.

Explanation:

Step1: Calculate $\Delta H^{\circ}_{r}$

The formula for $\Delta H^{\circ}_{r}$ is $\Delta H^{\circ}_{r}=Ea_{(fwd)}-Ea_{(rev)}$.
Substitute $Ea_{(fwd)} = + 65.0\ kJ$ and $Ea_{(rev)}=+150\ kJ$ into the formula:
$$\Delta H^{\circ}_{r}=65.0 - 150=-85.0\ kJ$$

Step2: Draw the potential - energy diagram

• Axes: The $x$ - axis represents the reaction progress, and the $y$ - axis represents the potential energy.
• Reactants: Label the potential energy level of reactants ($C + D$).
• Activated complex: Draw a peak above the reactants. The energy difference between reactants and the activated complex is $Ea_{(fwd)}=65.0\ kJ$.
• Products: The potential energy level of products ($CD$) is lower than that of reactants. The energy difference between the activated complex and products is $Ea_{(rev)} = 150\ kJ$. The energy difference between reactants and products is $\Delta H^{\circ}_{r}=-85.0\ kJ$.

Answer:

$\Delta H^{\circ}_{r}=-85.0\ kJ$