the value of δh° for the reaction below is -186 kj. h 2 (g) cl 2 (g) → 2hcl (g) the value of δh° f for hcl (g) is __ kj/mol.
The Correct Answer and Explanation is:
To find the value of ( \Delta H_f^\circ ) (standard enthalpy of formation) for HCl(g), we can use the relationship between the enthalpy change of a reaction and the standard enthalpies of formation of the reactants and products.
The given reaction is:
[
H_2(g) + Cl_2(g) \rightarrow 2HCl(g)
]
The standard enthalpy change of the reaction, ( \Delta H^\circ ), is -186 kJ. The standard enthalpy of formation, ( \Delta H_f^\circ ), is the enthalpy change when one mole of a compound is formed from its elements in their standard states. For elements like hydrogen and chlorine in their gaseous forms (( H_2(g) ) and ( Cl_2(g) )), the standard enthalpy of formation is zero, since they are in their most stable forms.
To find the ( \Delta H_f^\circ ) for HCl(g), we use the following equation for the reaction:
[
\Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) – \sum \Delta H_f^\circ (\text{reactants})
]
For the given reaction:
[
\Delta H^\circ = [2 \times \Delta H_f^\circ (\text{HCl(g)})] – [\Delta H_f^\circ (H_2(g)) + \Delta H_f^\circ (Cl_2(g))]
]
Since the standard enthalpy of formation of ( H_2(g) ) and ( Cl_2(g) ) is zero:
[
\Delta H^\circ = 2 \times \Delta H_f^\circ (\text{HCl(g)})
]
Substituting the known value for ( \Delta H^\circ ):
[
-186 \, \text{kJ} = 2 \times \Delta H_f^\circ (\text{HCl(g)})
]
Solving for ( \Delta H_f^\circ (\text{HCl(g)}) ):
[
\Delta H_f^\circ (\text{HCl(g)}) = \frac{-186}{2} = -93 \, \text{kJ/mol}
]
Final Answer:
The value of ( \Delta H_f^\circ ) for HCl(g) is -93 kJ/mol.
Explanation:
- The enthalpy of formation ( \Delta H_f^\circ ) refers to the heat change that occurs when one mole of a substance is formed from its constituent elements in their standard states.
- In this case, the elements hydrogen (H(_2)) and chlorine (Cl(_2)) are in their standard forms, so their ( \Delta H_f^\circ ) values are zero.
- The standard enthalpy change of the reaction ( \Delta H^\circ ) is given as -186 kJ, which is the heat released when hydrogen and chlorine gases react to form HCl gas.
- By using the enthalpy equation for reactions and knowing that the standard enthalpies of formation for H(_2(g)) and Cl(_2(g)) are zero, we can calculate the enthalpy of formation for HCl(g) as -93 kJ/mol.