Question

question 4
rustenburg/september 2024
the balanced equation below represents a hypothetical reaction:
x(g) + y(g) → xy(g)
the table below shows the different energies for the above reaction.

energy of the reactants (e_r)	351 kj.mol⁻¹
energy of the products (e_p)	109.2 kj.mol⁻¹
activation energy (e_a)	1370 kj.mol⁻¹

4.1 is the above reaction exothermic or endothermic? give a reason for the answer by referring to data in the table above.
4.2 calculate the heat of the above reaction.
4.3 define the term activated complex.
4.4 sketch the potential energy vs course of a reaction graph for the above reaction. the graph need not be drawn to scale. clearly label the axes and indicate the following:

• energy of the reactants
• energy of the products
• energy of the activated complex

4.5 what effect will the addition of a suitable catalyst to the reaction mixture have on the activation energy of the reaction? choose from increases, decreases or no effect.
4.6 the following reaction now takes place
xy(g) → x(g) + y(g)
for the above reaction, write down the value of:
4.6.1 heat of the reaction
4.6.2 activation energy

Explanation:

Step1: Determine reaction type

If $E_r>E_p$, exothermic; if $E_r < E_p$, endothermic. Given $E_r = 351\ kJ.mol^{-1}$ and $E_p=109.2\ kJ.mol^{-1}$, since $E_r>E_p$, the reaction is exothermic.

Step2: Calculate heat of reaction

The heat of reaction $\Delta H=E_p - E_r$. So $\Delta H=109.2 - 351=- 241.8\ kJ.mol^{-1}$.

Step3: Define activated complex

The activated complex is the unstable, high - energy intermediate state that reactant molecules form as they transform into products during a chemical reaction. It has the highest energy on the potential - energy diagram of the reaction.

Step4: Sketch potential energy graph

On the x - axis, label "Course of Reaction" and on the y - axis, label "Potential Energy". Mark a horizontal line for the energy of the reactants ($E_r = 351\ kJ.mol^{-1}$), a lower horizontal line for the energy of the products ($E_p = 109.2\ kJ.mol^{-1}$), and a peak for the energy of the activated complex ($E_a=1370\ kJ.mol^{-1}$). Draw a curve starting from the reactant energy level, rising to the activated - complex energy level and then falling to the product energy level.

Step5: Effect of catalyst on activation energy

A catalyst provides an alternative reaction pathway with a lower activation energy. So the addition of a suitable catalyst to the reaction mixture decreases the activation energy.

Step6: Determine values for reverse reaction

For the reverse reaction $XY(g)\to X(g)+Y(g)$, the heat of reaction is the negative of the heat of the forward reaction. So $\Delta H = 241.8\ kJ.mol^{-1}$. The activation energy of the reverse reaction is $E_a=1370 - 109.2 = 1260.8\ kJ.mol^{-1}$.

Answer:

4.1 The reaction is EXOTHERMIC because the energy of the reactants ($351\ kJ.mol^{-1}$) is greater than the energy of the products ($109.2\ kJ.mol^{-1}$).
4.2 $\Delta H=-241.8\ kJ.mol^{-1}$
4.3 The activated complex is the unstable, high - energy intermediate state that reactant molecules form as they transform into products during a chemical reaction.
4.4 (Sketch as described above)
4.5 DECREASES
4.6.1 $241.8\ kJ.mol^{-1}$
4.6.2 $1260.8\ kJ.mol^{-1}$