Answer :
Let's solve the problem step by step:
The given reaction is:
[tex]\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl(g) \][/tex]
We are provided with the enthalpy change of the reaction, [tex]\(\Delta H^{\circ}\)[/tex], which is -186 kJ.
[tex]\[ \Delta H^{\circ} = -186 \text{ kJ} \][/tex]
This enthalpy change of -186 kJ is for the entire reaction as written, which produces 2 moles of HCl gas.
To find the standard enthalpy of formation ([tex]\(\Delta H_f^{\circ}\)[/tex]) for [tex]\(HCl(g)\)[/tex], we need to determine the enthalpy change per mole of [tex]\(HCl\)[/tex].
Since the reaction produces 2 moles of [tex]\(HCl(g)\)[/tex]:
[tex]\[ \Delta H_f^{\circ} (HCl) = \frac{\Delta H^{\circ}}{\text{Number of moles of } HCl} \][/tex]
Plugging in the given enthalpy change and the number of moles:
[tex]\[ \Delta H_f^{\circ} (HCl) = \frac{-186 \text{ kJ}}{2 \text{ moles}} \][/tex]
[tex]\[ \Delta H_f^{\circ} (HCl) = -93 \text{ kJ/mol} \][/tex]
Thus, the value of [tex]\(\Delta H_f^{\circ}\)[/tex] for [tex]\(HCl(g)\)[/tex] is [tex]\(-93.0\)[/tex] kJ/mol.
Therefore, the correct answer is:
C) [tex]\(-93.0\)[/tex] kJ/mol
The given reaction is:
[tex]\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl(g) \][/tex]
We are provided with the enthalpy change of the reaction, [tex]\(\Delta H^{\circ}\)[/tex], which is -186 kJ.
[tex]\[ \Delta H^{\circ} = -186 \text{ kJ} \][/tex]
This enthalpy change of -186 kJ is for the entire reaction as written, which produces 2 moles of HCl gas.
To find the standard enthalpy of formation ([tex]\(\Delta H_f^{\circ}\)[/tex]) for [tex]\(HCl(g)\)[/tex], we need to determine the enthalpy change per mole of [tex]\(HCl\)[/tex].
Since the reaction produces 2 moles of [tex]\(HCl(g)\)[/tex]:
[tex]\[ \Delta H_f^{\circ} (HCl) = \frac{\Delta H^{\circ}}{\text{Number of moles of } HCl} \][/tex]
Plugging in the given enthalpy change and the number of moles:
[tex]\[ \Delta H_f^{\circ} (HCl) = \frac{-186 \text{ kJ}}{2 \text{ moles}} \][/tex]
[tex]\[ \Delta H_f^{\circ} (HCl) = -93 \text{ kJ/mol} \][/tex]
Thus, the value of [tex]\(\Delta H_f^{\circ}\)[/tex] for [tex]\(HCl(g)\)[/tex] is [tex]\(-93.0\)[/tex] kJ/mol.
Therefore, the correct answer is:
C) [tex]\(-93.0\)[/tex] kJ/mol
