A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)
2 H2O(l) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of HCl(aq) is kJ/mol.

Respuesta :

Answer: [tex]\Delta H^{0}=-173.72[/tex] kJ/mol

Explanation: Enthalpy Change is the amount of energy in a reaction - absorption or release - at a constant pressure. So, Standard Enthalpy of Formation is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

[tex]\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}[/tex]

[tex]Ca(OH)_{2}_{(aq)}+2HCl_{(aq)}[/tex] → [tex]CaCl_{2}_{(s)}+2H_{2}O_{(l)}[/tex]

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: [tex]\Delta H^{0}=[/tex] -1002.82 kJ/mol

Calcium chloride: [tex]\Delta H^{0}=[/tex] -795.8 kJ/mol

Water: [tex]\Delta H^{0}=[/tex] -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

[tex]-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H][/tex]

[tex]-17.2=-1367.46+1002.82-2\Delta H[/tex]

[tex]2\Delta H=17.2-364.64[/tex]

[tex]\Delta H=-173.72[/tex]

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

ACCESS MORE
EDU ACCESS
Universidad de Mexico