Question

Standard Gibbs free energy for the reaction X+Y=Z is -4.606Kcal The equilibrium constant for this reaction at 227degree celsius is

Answer 1

Delta G =-2.303 RT log k
- 4.606 ×10^3= - 2.303 * 8.134 * 500 log k
- 4.606 ×10^3/ - 9.573 ×10^3 =log k
log k= 0.4796
K=10^0.4796
K = anti log (0.4796)
K=3.017
Hope it helped you

Answer 2

Equilibrium constant for this reaction at 227 degree celsius is 26903

Explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = Standard Gibbs free energy = -4.606 kcal = 19271 Joules      (1kcal=4184 J)

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

T = temperature = 227 K

Putting values in above equation, we get:

$-19271J/mol=-(8.314J/Kmol)\times 227K\times \ln (K)$

$ln(K)=10.2$

$K=26903$

Hence, the equilibrium constant for this reaction at 227 degree celsius is 26903

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