Consider the reaction mg(oh)2(s)→mgo(s)+h2o(l) with enthalpy of reaction δhrxn∘=37.5kj/mol what is the enthalpy of formation of mgo(s)?

Answer is: -601,2 kJ/mol
Chemical reaction: Mg(OH)₂ → MgO + H₂O.
ΔHrxn = 37,5 kJ/mol.
ΔHf(Mg(OH)₂) = −924,5 kJ/mol.
ΔHf(H₂O) = −285,8 kJ/mol.
ΔHrxn -enthalpy of reaction.
ΔHf - enthalpy of formation.
ΔHrxn=∑productsΔHf−∑reactantsΔHf.
ΔHf(MgO) = -924,5 kJ/mol - (-285,8 kJ/mol) + 37,5 kj/mol.
ΔHf(MgO) = -601,2 kJ/mol.

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priyankatutor priyankatutor · Answer 2 · 2021-10-19T13:54:00+02:00

The enthalpy of MgO formation is -601.2kJ/mol in the given reaction.

Enthalpy:

It is the measure of energy constant at constant pressure. It is derived as [tex]\bold{\Delta H}[/tex].It is calculated by the formula,

[tex]\rm \bold{ \Delta H_r_x_n = \sum \Delta H_f_, product- \sum \Delta H_f_, reactant }[/tex]

Given here,

[tex]\rm \bold{ \Delta H_r_x_n = 37.5 kJ/mol}[/tex]

[tex]\rm \bold{ \Delta H_f(H_2O) = -285.8 kJ/mol}[/tex]

The reaction is

[tex]\rm \bold {Mg(OH)_2}\rightarrow\bold{ MgO_(_s_)+H_2O_(_l_)}[/tex]

Enthalpy of MgO formation will be,

[tex]\rm \bold {\Delta H_f(MgO) = -924.5-(-285.8)+ 37.5}\\\\\rm \bold {\Delta H_f(MgO) =-601.2 kJ/mol}[/tex]

Hence we can conclude that the enthalpy of MgO formation is -601.2kJ/mol.

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