Question

hydrogen reacts with chlorine to form hydrogen chloride hcl (g), $\delta h_f = -92.3$ kj/mol according to the reaction.\
$\ce{h2 (g) + cl2 (g) \
ightarrow 2hcl (g)}$\
which statement is correct?\
use $\delta h_{\text{rxn}} = \sum (\delta h_{f, \text{products}}) - \sum (\delta h_{f, \text{reactants}})$.\
\
the enthalpy of the\
reaction is $-184.6$ kj,\
and the reaction is\
endothermic.\
\
the enthalpy of the\
reaction is $184.6$ kj, and\
the reaction is exothermic.\
\
the enthalpy of the\
reaction is $-184.6$ kj,\
and the reaction is\
exothermic.\
\
the enthalpy of the\
reaction is $184.6$ kj, and\
the reaction is

Explanation:

Step1: Recall standard enthalpy of formation

For elements in their standard state (like \(H_2(g)\) and \(Cl_2(g)\)), \(\Delta H_f = 0\) kJ/mol.

Step2: Apply the reaction enthalpy formula

The reaction is \(H_2(g)+Cl_2(g)
ightarrow 2HCl(g)\). Using \(\Delta H_{rxn}=\sum(\Delta H_{f,products})-\sum(\Delta H_{f,reactants})\):

• \(\sum(\Delta H_{f,products}) = 2\times\Delta H_f(HCl(g)) = 2\times(-92.3\space kJ/mol)\)
• \(\sum(\Delta H_{f,reactants})=\Delta H_f(H_2(g))+\Delta H_f(Cl_2(g)) = 0 + 0 = 0\)

So, \(\Delta H_{rxn}=2\times(-92.3)-0=-184.6\space kJ\).

Step3: Determine reaction type

A negative \(\Delta H_{rxn}\) means the reaction releases heat (exothermic).

Answer:

The enthalpy of the reaction is \(-184.6\space kJ\), and the reaction is exothermic. (Assuming the third option in the boxes is this statement, e.g., if the third box is "The enthalpy of the reaction is \(-184.6\space kJ\), and the reaction is exothermic", then the answer is that option.)