Answer :
The reaction given is:
[tex]\[ C (s) + O_2 (g) \rightarrow CO_2 (g) \][/tex]
To determine the enthalpy change ([tex]\(\Delta H\)[/tex]) for this reaction, we use the standard enthalpies of formation ([tex]\(\Delta H_f\)[/tex]) of the reactants and products.
The enthalpy change of the reaction can be calculated using:
[tex]\[ \Delta H_{rxn} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}}) \][/tex]
For this reaction:
- The enthalpy of formation of carbon dioxide ([tex]\( CO_2(g) \)[/tex]) is given as [tex]\(-393.5 \text{ kJ/mol} \)[/tex].
- The standard enthalpy of formation for elemental carbon ([tex]\( C(s) \)[/tex]) and elemental oxygen ([tex]\( O_2(g) \)[/tex]) in their standard states is [tex]\(0 \text{ kJ/mol} \)[/tex].
Thus,
[tex]\[ \sum (\Delta H_{f, \text{products}}) = \Delta H_f (CO_2 (g)) = -393.5 \text{ kJ/mol} \][/tex]
And,
[tex]\[ \sum (\Delta H_{f, \text{reactants}}) = \Delta H_f (C (s)) + \Delta H_f (O_2 (g)) = 0 + 0 = 0 \text{ kJ/mol} \][/tex]
Substituting these values into the equation for [tex]\(\Delta H_{rxn}\)[/tex]:
[tex]\[ \Delta H_{rxn} = (-393.5) - (0) = -393.5 \text{ kJ/mol} \][/tex]
Therefore, the enthalpy change of the reaction is:
[tex]\[ -393.5 \text{ kJ} \][/tex]
So, the enthalpy change of the reaction is [tex]\(-393.5 \text{ kJ} \)[/tex].
[tex]\[ C (s) + O_2 (g) \rightarrow CO_2 (g) \][/tex]
To determine the enthalpy change ([tex]\(\Delta H\)[/tex]) for this reaction, we use the standard enthalpies of formation ([tex]\(\Delta H_f\)[/tex]) of the reactants and products.
The enthalpy change of the reaction can be calculated using:
[tex]\[ \Delta H_{rxn} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}}) \][/tex]
For this reaction:
- The enthalpy of formation of carbon dioxide ([tex]\( CO_2(g) \)[/tex]) is given as [tex]\(-393.5 \text{ kJ/mol} \)[/tex].
- The standard enthalpy of formation for elemental carbon ([tex]\( C(s) \)[/tex]) and elemental oxygen ([tex]\( O_2(g) \)[/tex]) in their standard states is [tex]\(0 \text{ kJ/mol} \)[/tex].
Thus,
[tex]\[ \sum (\Delta H_{f, \text{products}}) = \Delta H_f (CO_2 (g)) = -393.5 \text{ kJ/mol} \][/tex]
And,
[tex]\[ \sum (\Delta H_{f, \text{reactants}}) = \Delta H_f (C (s)) + \Delta H_f (O_2 (g)) = 0 + 0 = 0 \text{ kJ/mol} \][/tex]
Substituting these values into the equation for [tex]\(\Delta H_{rxn}\)[/tex]:
[tex]\[ \Delta H_{rxn} = (-393.5) - (0) = -393.5 \text{ kJ/mol} \][/tex]
Therefore, the enthalpy change of the reaction is:
[tex]\[ -393.5 \text{ kJ} \][/tex]
So, the enthalpy change of the reaction is [tex]\(-393.5 \text{ kJ} \)[/tex].
