Chemistry, 17.10.2019 04:30 Jacobolobo7

Consider this combination reaction:
2mg(s)+o2(g)→2mgo(s) δh=−1204 kj
what is the enthalpy for the decomposition of 1 mole of mgo(s) into mg(s) and o2(g)?
consider this combination reaction:
what is the enthalpy for the decomposition of 1 mole of into and ?
-1204 kj/mol
602 kj/mol
1204 kj/mol
-602 kj/mol

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Answer from: makayladurham19

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: link

2. Calculation of moles of HCl: link

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.

Answer from: Quest

No table found. the question asks for covalent bond based on electronegativies

Answer from: Quest

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