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

Answer:

The equilibrium constant for this reaction at 298.15 K is $2.067\times 10^{13}$.

Explanation:

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

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

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

$\Delta G^o_{rxn}=(2\times \Delta G^o_f_{(H_2O(g))}+2\times G^o_f_{(Cl_2(g))})-(4\times \Delta G^o_f_{(HCl(g))}+1\times G^o_f_{(O_2(g))})$

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:

$\Delta G^o_{rxn}=(2\times (-228.6 kJ/mol)+2\times 0 kJ/mol)-(4\times -95.3 kJ/mol+1\times 0 kJ/mol)=-76 kJ/mol$

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 = ?

$-76000 J/mol=-8.314J/K mol\times 298.145 K\ln K_1$

$\ln K_1=\frac{-76000 J/mol}{-8.314J/K mol\times 298.15 K}$

$K_1=2.067\times 10^{13}$

The equilibrium constant for this reaction at 298.15 K is $2.067\times 10^{13}$.