Questions: Hydrogen reacts with chlorine to form hydrogen chloride (HCl(g), ΔHf=-92.3 kJ/mol) according to the reaction below. H2(g) + Cl2(g) → 2 HCl(g) Which statement is correct? Use ΔHn×n=∑(ΔHt,products)−∑(ΔHt,reactants). 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 endothermic. 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.

Hydrogen reacts with chlorine to form hydrogen chloride (HCl(g), ΔHf=-92.3 kJ/mol) according to the reaction below.
H2(g) + Cl2(g) → 2 HCl(g)

Which statement is correct?
Use ΔHn×n=∑(ΔHt,products)−∑(ΔHt,reactants).
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 endothermic.
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.

Transcript text: Hydrogen reacts with chlorine to form hydrogen chloride $\left(\mathrm{HCl}(\mathrm{g}), \Delta H_{\mathrm{f}}=-92.3 \mathrm{~kJ} / \mathrm{mol}\right)$ according to the reaction below. \[ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{~g}) \] Which statement is correct? Use $\Delta H_{n \times n}=\sum\left(\Delta H_{\text {t,poducts }}\right)-\sum\left(\Delta H_{t, \text { reactants }}\right)$. 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 endothermic. 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.

Solution

Solution Steps

Step 1: Identify the Reaction Enthalpy Formula

The enthalpy change for a reaction ($\Delta H_{\text{reaction}}$) is calculated using the formula: \[ \Delta H_{\text{reaction}} = \sum \Delta H_{\text{products}} - \sum \Delta H_{\text{reactants}} \]

Step 2: Calculate the Enthalpy Change for Products

For the reaction $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{~g})$, the enthalpy of formation for $\mathrm{HCl}$ is given as $-92.3 \, \text{kJ/mol}$. Since there are 2 moles of $\mathrm{HCl}$ produced, the total enthalpy change for the products is: \[ 2 \times (-92.3 \, \text{kJ/mol}) = -184.6 \, \text{kJ} \]

Step 3: Calculate the Enthalpy Change for Reactants

The enthalpy of formation for elements in their standard state, such as $\mathrm{H}_{2}(\mathrm{~g})$ and $\mathrm{Cl}_{2}(\mathrm{~g})$, is zero. Therefore, the total enthalpy change for the reactants is: \[ 0 \, \text{kJ} \]

Step 4: Determine the Enthalpy Change of the Reaction

Using the formula from Step 1: \[ \Delta H_{\text{reaction}} = (-184.6 \, \text{kJ}) - (0 \, \text{kJ}) = -184.6 \, \text{kJ} \]

Step 5: Determine the Nature of the Reaction

Since the enthalpy change is negative ($-184.6 \, \text{kJ}$), the reaction is exothermic.

Step 6: Conclusion

The correct statement is: "The enthalpy of the reaction is -184.6 kJ, and the reaction is exothermic."