hydrogen reacts with chlorine to form hydrogen chloride (hcl (g), $delta h_f=-92.3$ kj/mol) according to the…

hydrogen reacts with chlorine to form hydrogen chloride (hcl (g), $delta h_f=-92.3$ kj/mol) according to the reaction below.\n$h_2(g)+cl_2(g)\rightarrow2hcl(g)$\nwhich statement is correct?\nuse $delta h_{rxn}=sum(delta h_{f,products})-sum(delta h_{f,reactants})$.\no the enthalpy of the reaction is -184.6 kj, and the reaction is exothermic.\no the enthalpy of the reaction is -184.6 kj, and the reaction is endothermic.\no the enthalpy of the reaction is 184.6 kj, and the reaction is endothermic.\no the enthalpy of the reaction is 184.6 kj, and the reaction is exothermic.

hydrogen reacts with chlorine to form hydrogen chloride (hcl (g), $delta h_f=-92.3$ kj/mol) according to the reaction below.\n$h_2(g)+cl_2(g)\rightarrow2hcl(g)$\nwhich statement is correct?\nuse $delta h_{rxn}=sum(delta h_{f,products})-sum(delta h_{f,reactants})$.\no the enthalpy of the reaction is -184.6 kj, and the reaction is exothermic.\no the enthalpy of the reaction is -184.6 kj, and the reaction is endothermic.\no the enthalpy of the reaction is 184.6 kj, and the reaction is endothermic.\no the enthalpy of the reaction is 184.6 kj, and the reaction is exothermic.

Answer

Explanation:

Step1: Identify reactants and products

Reactants: $H_2(g)$ and $Cl_2(g)$ with $\Delta H_f = 0$ (standard state elements). Product: $2HCl(g)$ with $\Delta H_f=- 92.3$ kJ/mol.

Step2: Calculate $\Delta H_{rxn}$

Using $\Delta H_{rxn}=\sum(\Delta H_{f,products})-\sum(\Delta H_{f,reactants})$, we have $\Delta H_{rxn} = 2\times(-92.3)- (0 + 0)=-184.6$ kJ.

Step3: Determine reaction type

Since $\Delta H_{rxn}<0$, heat is released, so it's an exothermic reaction.

Answer:

The enthalpy of the reaction is -184.6 kJ, and the reaction is exothermic.