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2 years ago

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: 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:

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

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

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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