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ankoles [38]
3 years ago
9

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
Chemistry
1 answer:
Ganezh [65]3 years ago
3 0

Answer:

ΔH °fCaO= -655.09 KJ/mol =  ΔH °fCaO

Explanation:

ΔH °(reaction)a=ΔH °f(product)-ΔH °f(reactant)

where,

ΔH °f(product)=is the standard heat of formation of products

-ΔH °f(reactant)=is the standard heat of formation of reactants

The standard heat of formation of  H 2 O ( l )  is  − 285.8 k J / m o l .

The standard heat of formation of  Ca(OH )2  is  − 986.09 k J / m o l .

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

ΔH °(reaction)a=ΔH °f(CaO+H2O)-ΔH °f(Ca(OH)₂

65.2                =  ΔH °fCaO+( -265.8) -(-986.09)

                       = ΔH °fCaO-265.8+986.09

65.2                        =  ΔH °fCaO+ 720.29

65.2-720.29      =  ΔH °fCaO

-655.09 KJ/mol ==  ΔH °fCaO

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