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jekas [21]
4 years ago
14

Consider this combination reaction:

Chemistry
2 answers:
Yuki888 [10]4 years ago
7 0
<h2>The molar enthalpy of decomposition of magnesium oxide is 602 kJ.</h2><h3>FURTHER EXPLANATION</h3><h3><u>Enthalpy is an Extensive Property</u></h3>

Enthalpy is the amount of heat in a system at constant pressure. It is an extensive property. Extensive properties are characteristics that are affected by the amount of substance. Examples of extensive properties are mass and length. The opposite of extensive properties are intensive properties. These are properties that are unique to a substance. They do not vary with the amount. Examples are density and boiling point.

Equations with enthalpy values are called thermochemical equations. The value of the ΔH is specific to the stoichiometric ratio provided in the equation. Therefore, the ΔH value of -1204 kJ is specific for the formation of 2 moles of magnesium oxide. The molar ΔH of formation should then be half of this which is -602 kJ.

<h3><u>Enthalpy depends on the direction of the reaction.</u></h3>

The direction of the reaction influences the enthalpy of the reaction. If the reaction is reversed, the enthalpy of the reversed reaction is equal in magnitude but opposite the sign of the forward reaction.

In this case, the reverse reaction of the formation of magnesium oxide is its decomposition. Therefore, the decomposition of one mole of magnesium oxide will have an enthalpy of +602 kJ.

<h3>LEARN MORE</h3>
  1. Learn more about Molar Enthalpy of Combustion brainly.com/question/10936616
  2. Learn more about Hess' Law brainly.com/question/3194633
  3. Learn more about Specific Heat Capacity brainly.com/question/12976104

Keywords: Enthalpy of formation, enthalpy of decomposition

amm18124 years ago
5 0

The enthalpy for the decomposition of one mole of {\text{MgO}} is  \boxed{602\;{\text{kJ}}}.

Further Explanation:

This problem is based upon Hess’s Law. Hess’s law is utilized for calculating the enthalpy change of a reaction that can be obtained simply by summation of two or more reactions. In accordance with the Hess’s law, \Delta H of an overall reaction is obtained by adding the enthalpy change for each individual step reaction involved to obtain the overall reaction.

\boxed{\Delta\text{H}_{\text{overallrxn}}=\Delta\text{H}_{1}+\Delta\text{H}_{2}+......+\Delta\text{H}_{n}}

Enthalpy is defined as state function and therefore its value depends upon the initial and final state of system but not upon the path. This is the reason that the overall reaction can be simply obtained by adding or subtracting the enthalpy change of the individual steps utilized to get the final reaction.

1. Combination reactions:

These reactions are also known as synthesis reaction. These are the reaction in which two or more reactants combine to form single product. These are generally accompanied by the release of heat so they are exothermic reactions.

Examples of combination reactions are as follows:

(a) {\text{Ba}} + {{\text{F}}_2}\to{\text{Ba}}{{\text{F}}_2}

(b) {\text{CaO}}+{{\text{H}}_2}{\text{O}} \to{\text{Ca}}{\left( {{\text{OH}}} \right)_2}

2. Decomposition reactions:

The opposite of combination reactions is called as decomposition reaction. Here, a single reactant gets broken into two or more products. Such reactions are usually endothermic because energy is required to break the existing bonds between the reactant molecules.

Examples of decomposition reactions are as follows:

(a) 2{{\text{H}}_2}{{\text{O}}_2}\to2{{\text{H}}_2}{\text{O}}+{{\text{O}}_2}

(b) 2{\text{NaCl}}\to{\text{2Na+C}}{{\text{l}}_2}

The combination reaction for the formation of {\text{MgO}} is as follows:

{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)\to2{\text{MgO}}\left(s\right)

The value of \Delta {H_1} is 180.7{\text{ kJ}}.

Step 1: The enthalpy change of the following reaction is \Delta {H_1} .

{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)\to2{\text{MgO}}\left(s\right)    ......(1)

The decomposition reaction of {\text{MgO}} is as follows:

2{\text{MgO}}\left(s\right)\to{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)

Step 2: The enthalpy change of the following reaction is \Delta {H_2}.

2{\text{MgO}}\left(s\right) \to{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)                                              ......(2)

The reaction (2) can be obtained by reversing the reaction (1) so the value of \Delta {H_2} can be obtained as follows:

\begin{aligned}\Delta {H_2}&=-\Delta {H_1}\\&=-\left({-1204\;{\text{kJ}}}\right)\\&=1204\;{\text{kJ}}\\\end{aligned}

In the decomposition reaction,two moles of {\mathbf{MgO}}  dissociates to give two moles of {\mathbf{Mg}} and one mole of {{\mathbf{O}}_{\mathbf{2}}} and therefore the enthalpy for the decomposition of one mole of  is as follows:

\begin{aligned}{\text{Enthalpy for decomposition of 1 mole}}&=\frac{{\Delta {H_2}}}{2}\\&=\frac{{1204\;{\text{kJ}}}}{2}\\&=602\;{\text{kJ}}\\\end{aligned}

Hence, enthalpy for decomposition of one mole of {\mathbf{MgO}} is {\mathbf{602}}\;{\mathbf{kJ}}.

Learn more:

1. Dissociation of ionic compounds: brainly.com/question/5425813

2. Calculation of moles of HCl: brainly.com/question/5950133

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Thermodynamics

Keywords: Hess’s Law, enthalpy, MgO, O2, Mg, 1204 kj, -1204 kj, 602 kj , -602 kj overall reaction, adding, state function, initial state, and final state.

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