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Sagot :
Let's analyze the given problem step by step based on the provided information.
1. Balanced Chemical Equation:
The balanced chemical equation is:
[tex]\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl(g) \][/tex]
2. Enthalpy of Formation:
The enthalpy of formation of hydrogen chloride (HCl) gas is given as:
[tex]\[ \Delta H_f = -92.3 \, \text{kJ/mol} \][/tex]
3. Number of Moles:
From the balanced chemical equation, we can see that 2 moles of HCl are produced.
4. Calculation of the Enthalpy Change of the Reaction:
Using the formula to calculate the total enthalpy change:
[tex]\[ \Delta H_{\text{reaction}} = \sum \left(\Delta H_{i,\text{products}} \right) - \sum \left(\Delta H_{i,\text{reactants}} \right) \][/tex]
Since [tex]\( H_2 \)[/tex] and [tex]\( Cl_2 \)[/tex] are elements in their standard state, their enthalpies of formation are zero. Therefore, we only need to consider the products:
[tex]\[ \Delta H_{\text{reaction}} = 2 \times (-92.3 \, \text{kJ/mol}) = -184.6 \, \text{kJ} \][/tex]
5. Nature of the Reaction:
The enthalpy change of the reaction is negative ([tex]\( \Delta H_{\text{reaction}} = -184.6 \, \text{kJ} \)[/tex]). A negative enthalpy change indicates that the reaction releases energy to the surroundings, thus it is exothermic.
Summarizing our findings:
- The enthalpy of the reaction is [tex]\(-184.6 \, \text{kJ}\)[/tex].
- The reaction is exothermic because it releases heat energy.
Based on this analysis, the correct statement is:
- The enthalpy of the reaction is [tex]\(-184.6 \, \text{kJ}\)[/tex], and the reaction is exothermic.
Therefore, the correct statement is:
The enthalpy of the reaction is -184.6 kJ, and the reaction is exothermic.
1. Balanced Chemical Equation:
The balanced chemical equation is:
[tex]\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl(g) \][/tex]
2. Enthalpy of Formation:
The enthalpy of formation of hydrogen chloride (HCl) gas is given as:
[tex]\[ \Delta H_f = -92.3 \, \text{kJ/mol} \][/tex]
3. Number of Moles:
From the balanced chemical equation, we can see that 2 moles of HCl are produced.
4. Calculation of the Enthalpy Change of the Reaction:
Using the formula to calculate the total enthalpy change:
[tex]\[ \Delta H_{\text{reaction}} = \sum \left(\Delta H_{i,\text{products}} \right) - \sum \left(\Delta H_{i,\text{reactants}} \right) \][/tex]
Since [tex]\( H_2 \)[/tex] and [tex]\( Cl_2 \)[/tex] are elements in their standard state, their enthalpies of formation are zero. Therefore, we only need to consider the products:
[tex]\[ \Delta H_{\text{reaction}} = 2 \times (-92.3 \, \text{kJ/mol}) = -184.6 \, \text{kJ} \][/tex]
5. Nature of the Reaction:
The enthalpy change of the reaction is negative ([tex]\( \Delta H_{\text{reaction}} = -184.6 \, \text{kJ} \)[/tex]). A negative enthalpy change indicates that the reaction releases energy to the surroundings, thus it is exothermic.
Summarizing our findings:
- The enthalpy of the reaction is [tex]\(-184.6 \, \text{kJ}\)[/tex].
- The reaction is exothermic because it releases heat energy.
Based on this analysis, the correct statement is:
- The enthalpy of the reaction is [tex]\(-184.6 \, \text{kJ}\)[/tex], and the reaction is exothermic.
Therefore, the correct statement is:
The enthalpy of the reaction is -184.6 kJ, and the reaction is exothermic.
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