Consider the reaction Ca(OH)2(s)→CaO(s)+H2O(l) with enthalpy of reaction ΔHrxn∘=65.2kJ/mol What is the enthalpy of formation of CaO(s)? Express your answer in kilojoules per mole to one deci

Enthalpy is defined as the heat content of reactants or products under constant pressure. The enthalpy change of the reaction can be calculated by

[tex]\Delta \text H_{\text reaction} ^{o} &=& \text H_{\text product}^{o} - \text H_{\text reactant}^{o}[/tex]. In the given question, enthalpy of formation of calcium oxide is (CaO) -635.1kJ/mol.

Explanation:

In the reaction, it is given that calcium hydroxide is formed by the reaction of calcium oxide and water. The equation can be represented as:

[tex]\text Ca{\text {(OH})_{2} \rightarrow\; \text {CaO} + \text {H}_{2} \text {O} }[/tex]

The equation of enthalpy is written as:

[tex]\Delta \text H_{\text reaction} ^{o} &=& \text H_{\text {product}}^{o} - \text H_{\text reactant}^{o}[/tex]

in which,

[tex]\Delta \text H_{\text {reactant} ^{o} &=&[/tex] standard heat of formation of reactants

[tex]\Delta \text H_{\text {product} ^{o} &=&[/tex] standard heat of formation of products

As we know,

Standard heat of formation of water is -285.8 kJ/ mol.

Standard heat of formation of calcium hydroxide is -986.1 kJ/mol.

From the equation of standard enthalpy change, it can be calculated as:

[tex]\Delta \text H_{\text Ca (OH)_{2}} ^{o} &=& ({\text {H} _{\text CaO} ^{o} + {\text {H} _{\text H_{2}O } ^{o})- \text H_{\text Ca (OH)_{2}}^{o}[/tex]

Substituting the values,

[tex]\begin{aligned}\text 65.2} ^{o} &={\text {H} _{\text CaO} ^{o} + (-285.8) - (-986.1) \\\text {H} _{\text CaO} ^{o} &= -635.1 \text {kJ/mol}\end {aligned}[/tex]

Thus, the given reaction is an exothermic reaction, which is denoted by the negative sign.

For Further Reference:

https://brainly.com/question/23729946?referrer=searchResults