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tatuchka [14]
2 years ago
5

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)

Chemistry
1 answer:
elixir [45]2 years ago
8 0

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

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

The standard enthalpy of formation of HCl is calculated as:

\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}

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

Standard Enthalpy of formation for the other compounds are:

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

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

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

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

Calculating:

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

-17.2=-1367.46+1002.82-2\Delta H

2\Delta H=17.2-364.64

\Delta H=-173.72

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

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