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Sagot :
Certainly! Let's analyze the given reaction and determine the enthalpy change and whether the reaction is exothermic or endothermic.
The balanced chemical reaction is:
[tex]\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \][/tex]
1. Enthalpy Change for the Products:
- Given: The enthalpy change of formation ([tex]\(\Delta H_f\)[/tex]) for HCl is [tex]\(-92.3 \text{ kJ/mol}\)[/tex].
- Since 2 moles of HCl are produced, the total enthalpy change for the products will be:
[tex]\[ \Delta H_{\text{products}} = 2 \times (-92.3 \text{ kJ/mol}) = -184.6 \text{ kJ} \][/tex]
2. Enthalpy Change for the Reactants:
- Both [tex]\(\text{H}_2(g)\)[/tex] and [tex]\(\text{Cl}_2(g)\)[/tex] are in their standard states, so their enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) is 0.
- Therefore, the total enthalpy change for the reactants is:
[tex]\[ \Delta H_{\text{reactants}} = 0 \text{ kJ} \][/tex]
3. Calculating the Enthalpy of the Reaction:
- According to the equation for the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \sum \left( \Delta H_{\text{f,products}} \right) - \sum \left( \Delta H_{\text{f,reactants}} \right) \][/tex]
- Substituting in the values we have:
[tex]\[ \Delta H_{\text{reaction}} = -184.6 \text{ kJ} - 0 \text{ kJ} = -184.6 \text{ kJ} \][/tex]
4. Determining if the Reaction is Exothermic or Endothermic:
- A reaction is exothermic if [tex]\(\Delta H_{\text{reaction}}\)[/tex] is negative.
- A reaction is endothermic if [tex]\(\Delta H_{\text{reaction}}\)[/tex] is positive.
- In this case, [tex]\(\Delta H_{\text{reaction}} = -184.6 \text{ kJ}\)[/tex] which is negative.
Therefore, the enthalpy of the reaction is [tex]\(-184.6 \text{ kJ}\)[/tex], and the reaction is exothermic.
The balanced chemical reaction is:
[tex]\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \][/tex]
1. Enthalpy Change for the Products:
- Given: The enthalpy change of formation ([tex]\(\Delta H_f\)[/tex]) for HCl is [tex]\(-92.3 \text{ kJ/mol}\)[/tex].
- Since 2 moles of HCl are produced, the total enthalpy change for the products will be:
[tex]\[ \Delta H_{\text{products}} = 2 \times (-92.3 \text{ kJ/mol}) = -184.6 \text{ kJ} \][/tex]
2. Enthalpy Change for the Reactants:
- Both [tex]\(\text{H}_2(g)\)[/tex] and [tex]\(\text{Cl}_2(g)\)[/tex] are in their standard states, so their enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) is 0.
- Therefore, the total enthalpy change for the reactants is:
[tex]\[ \Delta H_{\text{reactants}} = 0 \text{ kJ} \][/tex]
3. Calculating the Enthalpy of the Reaction:
- According to the equation for the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \sum \left( \Delta H_{\text{f,products}} \right) - \sum \left( \Delta H_{\text{f,reactants}} \right) \][/tex]
- Substituting in the values we have:
[tex]\[ \Delta H_{\text{reaction}} = -184.6 \text{ kJ} - 0 \text{ kJ} = -184.6 \text{ kJ} \][/tex]
4. Determining if the Reaction is Exothermic or Endothermic:
- A reaction is exothermic if [tex]\(\Delta H_{\text{reaction}}\)[/tex] is negative.
- A reaction is endothermic if [tex]\(\Delta H_{\text{reaction}}\)[/tex] is positive.
- In this case, [tex]\(\Delta H_{\text{reaction}} = -184.6 \text{ kJ}\)[/tex] which is negative.
Therefore, the enthalpy of the reaction is [tex]\(-184.6 \text{ kJ}\)[/tex], and the reaction is exothermic.
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