Question

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 1

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.

Answer 2

The enthalpy of the formation of MgO(s) is $\boxed{ - {\text{602 kJ}}}$

Further Explanation:

Enthalpy:

It is a thermodynamic property that is defined as the sum of internal energy and product of pressure (P) and volume (V) of the system. It is a state function, an extensive property, and is independent of the path followed by the system while moving from initial to the final point. The total enthalpy of the system cannot be measured directly so its change $\left({\Delta {\text{H}}}\right)$  is usually measured.

The enthalpy change $\left({\Delta {\text{H}}}\right)$ can have two values:

Case I: If the reaction is endothermic, more energy needs to be supplied to the system than that released by it. So $\Delta {\text{H}}$ comes out to be positive.

Case II: If the reaction is exothermic, more energy is released by the system than that supplied to it. So $\Delta {\text{H}}$ comes out to be negative.

Standard enthalpy of reaction:

This is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactants at the standard conditions. The expression to calculate the standard enthalpy of reaction $\left(\Delta H_{{\text{rxn}}}^{\circ}\right)$ is as follows:

$=\sum m\Delta H^{\circ}_{\text{f(products)}}-\sum n\Delta H^{\circ}_{\text{f(reactants)}}$

Here,

m is the stoichiometric coefficient of the product.

n is the stoichiometric coefficient of reactant.

$\Delta H_{\text{f(reactants)}}^{\circ}$ is the standard enthalpy of reactant formation.

$\Delta H_{\text{f(products)}}^{\circ}$  is the standard enthalpy of product formation.

The given reaction is as follows;

${\text{Mg}}{\left( {{\text{OH}}} \right)_2}\left(s\right)\to{\text{MgO}}\left( s \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)$

One mole of ${\text{Mg}}{\left( {{\text{OH}}}\right)_2}$  decomposes to form one mole of MgO and one mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ .

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{{\text{rxn}}}^{\circ}\right)$ is as follows:

$\Delta H_{rxn}^{\circ}=\left[\{\Delta H_{\text{f}}^{\circ}[\text{MgO(s)}]+\Delta H_{\text{f}}^{\circ}[\text{H}_{2}\text{O(l)}]\}-\{\Delta H_{\text{f}}^{\circ}[\text{Mg(OH)}_{2}\text{(s)}]\}\right]$           ......(1)

Rearrange equation (1) to calculate the standard enthalpy of formation of MgO.

$\Delta H_{\text{f}}^{\circ}[\text{MgO(s)}]=\Delta H_{rxn}^{\circ}+\Delta H_{\text{f}}^{\circ}[\text{Mg(OH)}_{2}\text{(s)}]-\Delta H_{\text{f}}^{\circ}[\text{H}_{2}\text{O(l)}]$

The value of $\left(\Delta H_{{\text{rxn}}}^{\circ}\right)$ is 37.5 kJ/mol.

The value of $\Delta H_{\text{f}}^{\circ}[\text{Mg(OH)}_{2}\text{(s)}]$ is -924.5 kJ/mol.

The value of $\Delta H_{\text{f}}^{\circ}[\text{H}_{2}\text{O(l)}]$ is -285.5 kJ/mol.

Substitute these values in equation (2).

$\begin{aligned}\Delta H_{\text{f}}^{\circ}[\text{Mg(OH)}_{2}\text{(s)}]&={\text{37}}{\text{.5 kJ/mol}}+\left({-{\text{925 kJ/mol}}}\right)-\left({-{\text{285}}{\text{.5 kJ/mol}}}\right)\\&=-60{\text{2 kJ/mol}}\end{aligned}$

Hence the enthalpy of formation of MgO is -602 kJ/mol.

Learn more:

1. Calculate the enthalpy change using Hess’s Law: brainly.com/question/11293201

2. Find the enthalpy of decomposition of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: enthalpy of formation, MgO, Mg(OH)2, H2O, -285.5 kJ/mol, 37.5 kJ/mol, -925 kJ/mol, enthalpy of reaction, -602 kJ/mol, intensive, extensive.