Question

What is the equilibrium constant for the reaction:

SO2 (g) + NO2 (g) â SO3 (g) + NO (g)

At 298 K? Use the following data: R=8.314 J/(K.mol)

Substance SO2 (g) SO3 (g) NO2 (g) NO (g)

ÎGo (kJ/mol) -300.2 -371 51 86.6

a) 6.8 . 10-7

b) 1.5 . 106

c) 1.014

d) 0.986

e) -35.2

Answer 1

The correct answer is option b.

Step-by-step explanation:

`SO_2 (g)+NO_2 (g)\rightarrow SO_3 (g)+NO (g)`

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

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

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

`\Delta G^o_(rxn)=(\Delta G^o_f_((SO_3(g)))+\Delta G^o_f_((NO(g))))-(\Delta G^o_f_((SO_2(s)))+G^o_f_((NO_2(g))))`

We are given:

`\Delta G^o_f_((SO_2(s)))=-300.2 kJ/mol\\\Delta G^o_f_((NO_2(g)))=51 kJ/mol`

`\Delta G^o_f_((SO_3(s)))=-371 kJ/mol\\\Delta G^o_f_((NO(g)))=86.6kJ/mol`

Putting values in above equation, we get:

`\Delta G^o_(rxn)=(-371 kJ/mol+86.6kJ/mol)-(-300.2 kJ/mol+51 kJ/mol)=-35.2 kJ/mol`

To calculate the
`K_c` (at 298 K) for given value of Gibbs free energy, we use the relation:

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

where,

`\Delta G^o` = Gibbs free energy = -35.2 kJ/mol = -35200 J/mol (Conversion factor: 1kJ = 1000J)

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

T =Temperature =298K[/tex]

`K_1` = equilibrium constant at 298 k;

Putting values in above equation, we get:

`-35200 J/mol=-(8.314J/Kmol)* 298K* \ln K_c\\\\K_c=1.479* 10^6\approx 1.5* 10^6`

Hence, correct answer is option b.