Question

data tables required:
note:
the enthalpy of formation of butane (c₄h₁₀) is not listed below,
but it is - 125.5 kj/mol
rates of reactions

1. consider the following reaction:

2ab₍g₎→2a₍g₎+b₍g₎
ea₍fwd₎=61.0 kj
ea₍rev₎=150.0 kj
a) calculate the enthalpy change (δhᵣ°) of this reaction
and determine if the reaction is endothermic or
exothermic in the forward direction.
b) draw and label a potential energy diagram including
your value for δhᵣ°.

Explanation:

Part (a)

Step 1: Recall the formula for enthalpy change

The enthalpy change of a reaction ($\Delta H^{\circ}$) is related to the activation energies of the forward ($Ea_{fwd}$) and reverse ($Ea_{rev}$) reactions by the formula: $\Delta H^{\circ} = Ea_{fwd} - Ea_{rev}$

Step 2: Substitute the given values

We know that $Ea_{fwd} = 61.0\ kJ$ and $Ea_{rev} = 150.0\ kJ$. Substituting these into the formula:
$\Delta H^{\circ} = 61.0\ kJ - 150.0\ kJ$

Step 3: Calculate the result

$61.0 - 150.0 = -89.0\ kJ$

Step 4: Determine if the reaction is endothermic or exothermic

If $\Delta H^{\circ}$ is negative, the reaction is exothermic (releases heat). If it is positive, the reaction is endothermic (absorbs heat). Since our calculated $\Delta H^{\circ} = -89.0\ kJ$ (negative), the reaction is exothermic in the forward direction.

1. Axes: Draw a vertical axis labeled "Potential Energy (kJ)" and a horizontal axis labeled "Reaction Progress".
2. Reactants and Products:

• Mark the potential energy of the reactants ($2AB_{(g)}$) at a certain level on the vertical axis.
• The activation energy for the forward reaction ($Ea_{fwd} = 61.0\ kJ$) is the energy difference between the reactants and the transition state. So, the transition state will be $61.0\ kJ$ above the reactants' potential energy.
• The activation energy for the reverse reaction ($Ea_{rev} = 150.0\ kJ$) is the energy difference between the products and the transition state. So, the products' potential energy will be $Ea_{rev}-Ea_{fwd}= 150.0 - 61.0=89.0\ kJ$ below the transition state (or since $\Delta H^{\circ}=- 89.0\ kJ$, the products will be $89.0\ kJ$ below the reactants in terms of potential energy).
• Label the reactants as $2AB_{(g)}$, the products as $2A_{(g)} + B_{(g)}$, the transition state, $Ea_{fwd}$, $Ea_{rev}$, and $\Delta H^{\circ}$ on the diagram.

Answer:

The enthalpy change ($\Delta H^{\circ}$) of the reaction is $\boldsymbol{-89.0\ kJ}$, and the reaction is exothermic in the forward direction.

Part (b)