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Skme · Answer 1 · 2023-01-31T19:43:36+0000

$C_2H_(4(g))+3O_(2(g))\rightarrow2CO_(2(g))+2H_2O_((g))$

We will use the standard-state energy of formation to calculate the reaction Gibbs free energy. We first have to calculate the entropy and the enthalppy of the reaction using the standard state conditions.

$\begin{gathered} \Delta H\degree=\Sigma\Delta H\degree_{f\text{ }products}-\Sigma\Delta H\degree_{f\text{ }reactants} \\ \Delta H\degree=(2*-393.5kJmol^(-1))+(2*-241.8kJmol^(-1))-(3*0)+(52.3kJmol^(-1)) \\ \Delta H\degree=-1,218.3\text{ }kJmol^(-1) \\ \\ \Delta S\operatorname{\degree}=\Sigma\Delta S\operatorname{\degree}_{f\text{p}roducts}-\Sigma\Delta S\operatorname{\degree}_{f\text{r}eactants} \\ \Delta S\operatorname{\degree}=(2*213.6Jmol^(-1)K^(-1))+(2*188.7Jmol^(-1)K^(-1))-(3*205.0Jmol^(-1)K^(-1)) \\ +(219.5Jmol^(-1)K^(-1)) \\ \Delta S\operatorname{\degree}=-29.9Jmol^(-1)K^(-1) \end{gathered}$

We will now substitute the values into the free energy equation to determine Gibbs energy for the reaction:

$\begin{gathered} \Delta G\degree=\Delta H\degree-T\Delta S\degree \\ \end{gathered}$

But first we need to convert the units for entropy S and the temeprature to Kelvin:

$\begin{gathered} \Delta S\degree=-29.9Jmol^(-1)K^(-1)*((1kJ)/(1000J)) \\ \Delta S\degree=-0.0299kJmol^(-1)K^(-1) \\ \\ T=273.15K+450\degree C \\ T=723.15K \end{gathered}$
$\begin{gathered} \Delta G\degree=-1,218.3\text{ }kJmol^(-1)-(723.15K*-0.0299kJmol^(-1)K^(-1)) \\ \Delta G\degree=-1,196.78kJmol^(-1) \end{gathered}$

Answer: The free-energy is -1,196kJmol^-1.

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