Question

Consider the reaction 4 HCl(g) + O2(g) =2 H2O(g) + 2 Cl2(g) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K. ANSWER:_______

delta G (kJ/mol)

HCL=-95.3

O2=0

H2O=-228.6

Cl2=0

Answer 1

The equilibrium constant for this reaction at 298.15 K is
`2.067* 10^(13)`.

Step-by-step explanation:

The equation used to calculate Gibbs free change is of a reaction is:

`\Delta G^o_(rxn)=\sum [n* \Delta G^o_f(product)]-\sum [n* \Delta G^o_f(reactant)]`

The equation for the enthalpy change of the above reaction is:

`\Delta G^o_(rxn)=(2* \Delta G^o_f_((H_2O(g)))+2* G^o_f_((Cl_2(g))))-(4* \Delta G^o_f_((HCl(g)))+1* G^o_f_((O_2(g))))`

We are given:

`\Delta G^o_f_((HCl(g)))=-95.3 kJ/mol\\\Delta G^o_f_((H_2O(g)))=228.6 kJ/mol`

`\Delta G^o_f_((O_2(g)))=\Delta G^o_f_((Cl_2(g)))=0` (pure element)

Putting values in above equation, we get:

`\Delta G^o_(rxn)=(2* (-228.6 kJ/mol)+2* 0 kJ/mol)-(4* -95.3 kJ/mol+1* 0 kJ/mol)=-76 kJ/mol`

To calculate the
`K_1` (at 25°C) for given value of Gibbs free energy, we use the relation:

`\Delta G^o=-RT\ln K_1`

where,

`\Delta G^o` = Gibbs free energy = -76 kJ/mol = -76000 J/mol

(Conversion factor: 1kJ = 1000J)

R = Gas constant =
`8.314J/K mol`

T = temperature = 298.15 K[/tex]

`K_1` = equilibrium constant at 25°C = ?

`-76000 J/mol=-8.314J/K mol* 298.145 K\ln K_1`

`\ln K_1=(-76000 J/mol)/(-8.314J/K mol* 298.15 K)`

`K_1=2.067* 10^(13)`

The equilibrium constant for this reaction at 298.15 K is
`2.067* 10^(13)`.