Question

Calculate the free-energy change of the following reaction at 450°C and standard pressure. Values in the table are at standard pressure and 25°C._______ kJ/mol

Answer 1

`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.

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.