Answer :
To address this problem, we need to determine the enthalpy change for the reaction given after it is reversed. The enthalpy change for the reaction is initially provided as [tex]\( \Delta H = -286 \)[/tex] kJ.
When a chemical reaction is reversed, the sign of the enthalpy change is also reversed. This is a fundamental concept in thermodynamics associated with Hess's Law, which allows us to manipulate reactions to find the enthalpy changes of other reactions. Here are the steps to determine the enthalpy of the reversed reaction:
1. Identify the given enthalpy change:
[tex]\[ \Delta H_{\text{forward}} = -286 \, \text{kJ} \][/tex]
2. Reverse the reaction:
The original reaction is:
[tex]\[ H_2 + 0.5 \, O_2 \rightarrow H_2O \][/tex]
The reversed reaction will be:
[tex]\[ H_2O \rightarrow H_2 + 0.5 \, O_2 \][/tex]
3. Change the sign of the enthalpy change:
When reversing the reaction, the enthalpy change will be the opposite in sign. Thus:
[tex]\[ \Delta H_{\text{reversed}} = -(-286 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{reversed}} = 286 \, \text{kJ} \][/tex]
Therefore, the final value for the enthalpy of the reversed intermediate reaction that you use will be [tex]\( 286 \, \text{kJ} \)[/tex].
The correct choice is:
[tex]\[ \text{A. } 286 \, \text{kJ} \][/tex]
When a chemical reaction is reversed, the sign of the enthalpy change is also reversed. This is a fundamental concept in thermodynamics associated with Hess's Law, which allows us to manipulate reactions to find the enthalpy changes of other reactions. Here are the steps to determine the enthalpy of the reversed reaction:
1. Identify the given enthalpy change:
[tex]\[ \Delta H_{\text{forward}} = -286 \, \text{kJ} \][/tex]
2. Reverse the reaction:
The original reaction is:
[tex]\[ H_2 + 0.5 \, O_2 \rightarrow H_2O \][/tex]
The reversed reaction will be:
[tex]\[ H_2O \rightarrow H_2 + 0.5 \, O_2 \][/tex]
3. Change the sign of the enthalpy change:
When reversing the reaction, the enthalpy change will be the opposite in sign. Thus:
[tex]\[ \Delta H_{\text{reversed}} = -(-286 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{reversed}} = 286 \, \text{kJ} \][/tex]
Therefore, the final value for the enthalpy of the reversed intermediate reaction that you use will be [tex]\( 286 \, \text{kJ} \)[/tex].
The correct choice is:
[tex]\[ \text{A. } 286 \, \text{kJ} \][/tex]
