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H 2(g)+I 2(g)→2 H I(g)
ΔH∘r x n=−9.5 k J/m o l r x n

H 2(g) and I 2(g) can react to produce HI(g) , as represented above. Shown below is an incomplete energy diagram for the reaction. The potential energy of the reactants is indicated on the diagram.

a) Carefully complete the energy diagram by drawing a curve that accurately shows the progress of the reaction, beginning at the reactants, moving through the transition state, and ending at the final products. When adding the products to the diagram, be sure to use an energy level that is appropriate relative to the energy level of the reactants
(b) On your energy diagram, draw a dashed curve to show the reaction progress from reactants to products in the presence of a suitable catalyst for the reaction.
(c) On your energy diagram, draw a vertical line segment with a length that corresponds to the activation energy for the reverse u n-catalyzed reaction:
2 HI(g)→H 2(g)+ I 2(g) .

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Answer:

The images for the 3 sub-questions have been presented in the first, second and third attached image to this solution.

Explanation:

The reaction for the reaction between Hydrogen gas and Iodine to give Hydrogen iodide

H₂ (g) + I₂ (g) → 2 H I(g)

ΔH∘rxn = −9.5 kJ/mol r x n

a) From the information provided, the heat of reaction being negative shows that the reaction is an exothermic reaction with the products for an exothermic reaction having a lower energy content/level as the reactants for this reaction.

This shows that the energy diagram will have the products at a lower level than fe reactants.

The sketch of the energy diagram for this exothermic reaction is presented in the first attached image to this solution.

The heat of reaction is shown on the energy diagram, together with the activation energy, Ea, the heat content/energy level for the reactants (Hr) & products (Hr) and finally the ΔH∘rxn.

ΔH∘rxn = Hp - Hr

b) When a catalyst that speed up a chemical reaction is introduced to the chemical reaction, the catalyst goes about making the reaction faster by lowering the activation energy of the reaction.

The activation energy is the minimum energy that the reactants must possess to the able to form products. On the graph, it is denoted as Ea, the difference in energy between the peak of the curve and the emergy level of the reactants.

So, a catalyst will cause this activation energy to be lowered to a new level of Ea₁.

The energy diagram of this is presented in the second attached image to this answer.

The ΔH∘rxn remains unchanged for this introduction of catalyst.

The broken lines are used to represent the lowered activation energy action of the catalyst.

c) The reverse reaction will have opposite the energy properties of the initial reaction.

The initially exothermic reaction with higher heat content of reactants than products is now an endothermic reaction with heat content/energy level of the products higher than that of the reactants.

The energy diagram is presented in the third attached image.

The activation energy is bigger and the ΔH∘rxn is now positive and equal to 9.5 kJ/mol, the direct additive inverse of the ΔH∘rxn for the initial exothermic reaction.

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Ver imagen AyBaba7
Ver imagen AyBaba7
Ver imagen AyBaba7

Catalyst increases the rate of reaction by decreasing the activation energy. The energy diagram will be reversed if the reaction is reversed.

Given here,

[tex]\bold {H_2 (g) + I_2 (g) \rightarrow 2 H I(g)}[/tex]

Enthalpy of the reaction

[tex]\bold {\Delta H = -9.5 kJ/mol}[/tex]

A) The sign is negative hence the reaction is exothermic. It means that the product will be lower on the energy diagram.

B) the catalyst reduces the activation energy of the reaction. Hence, on the addition of the catalyst the energy curve will go down.

C) The reversed reaction, the reaction becomes endothermic. Hence, the product contains more energy than the reactance.

Therefore the energy diagram will be reversed if the reaction reversed

To know more about the energy diagram, refer to the link:

https://brainly.com/question/12078312