Determine the standard enthalpy of formation for hydrogen chloride gas using the following data:H2(g) + Cl2(g) → 2HCl(g) ΔH = -184.6 kJ/molH2(g) + 1/2O2(g) → H2O(l) ΔH = -285.8 kJ/mol2HCl(g) + O2(g) → 2ClO(g) + H2O(l) ΔH = -518.2 kJ/mol Note: The standard enthalpy of formation for a compound is the change in enthalpy that occurs when one mole of the compound is formed from its constituent elements, with all substances in their standard state at a given temperature (usually 25°C) and pressure (usually 1 atm).

Question

Determine the standard enthalpy of formation for hydrogen chloride gas using the following data:H2(g) + Cl2(g) → 2HCl(g)    ΔH = -184.6 kJ/molH2(g) + 1/2O2(g) → H2O(l)  ΔH = -285.8 kJ/mol2HCl(g) + O2(g) → 2ClO(g) + H2O(l)   ΔH = -518.2 kJ/mol Note: The standard enthalpy of formation for a compound is the change in enthalpy that occurs when one mole of the compound is formed

Answer 1

To determine the standard enthalpy of formation for hydrogen chloride gas  HCl g  , we can use Hess's Law, which states that the total enthalpy change for a reaction is the same, regardless of the pathway taken. We can manipulate the given reactions to create a new reaction that represents the formation of HCl g  from its constituent elements.First, let's write down the target reaction we want to find the enthalpy change for:1/2H2 g  + 1/2Cl2 g   HCl g    H = ?Now, let's manipulate the given reactions to create the target reaction:1. H2 g  + Cl2 g   2HCl g     H = -184.6 kJ/molSince we want only 1 mole of HCl g  in the target reaction, we can divide this reaction by 2:1/2H2 g  + 1/2Cl2 g   HCl g    H = -92.3 kJ/mol2. H2 g  + 1/2O2 g   H2O l   H = -285.8 kJ/molWe don't need this reaction for the target reaction, so we can ignore it.3. 2HCl g  + O2 g   2ClO g  + H2O l    H = -518.2 kJ/molWe need to reverse this reaction and divide it by 2 to get the desired reactants and products:HCl g   1/2H2 g  + 1/2Cl2 g  + 1/2O2 g  - 1/2H2O l    H = 259.1 kJ/molNow, we can add the manipulated reactions to get the target reaction:1/2H2 g  + 1/2Cl2 g   HCl g    H = -92.3 kJ/molHCl g   1/2H2 g  + 1/2Cl2 g  + 1/2O2 g  - 1/2H2O l    H = 259.1 kJ/mol--------------------------------------------------------------1/2H2 g  + 1/2Cl2 g   HCl g    H = -92.3 + 259.1 = 166.8 kJ/molTherefore, the standard enthalpy of formation for hydrogen chloride gas  HCl g   is 166.8 kJ/mol.