Chemistry, 22.07.2019 05:00 terryonsavage543

Explain why the total amount of energy does not decrease in an exergonic chemical reaction

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An exergonic reaction is a chemical reaction where the change in the free energy is negative (there is a net release of free energy),[1] indicating a spontaneous reaction. For processes that take place under constant pressure and temperature conditions, the Gibbs free energy is used whereas the Helmholtz energy is used for processes that take place under constant volume and temperature conditions.

Symbolically, the release of free energy, G, in an exergonic reaction (at constant pressure and temperature) is denoted as

{\displaystyle \Delta G=G_{\rm {products}}-G_{\rm {reactants}}<0.\,}

Although exergonic reactions are said to occur spontaneously, this does not imply that the reaction will take place at an observable rate. For instance, the disproportionation of hydrogen peroxide is very slow in the absence of a suitable catalyst. It has been suggested that eager would be a more intuitive term in this context.[2]

More generally, the terms exergonic and endergonic relate to the free energy change in any process, not just chemical reactions. An example of an exergonic reaction is cellular respiration. This relates to the degrees of freedom as a consequence of entropy, the temperature, and the difference in heat released or absorbed.

By contrast, the terms exothermic and endothermic relate to the overall exchange of heat during a process

Answer from: Quest

explanation:

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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