Question

the standard free energy change for the decomposition of two moles of hydrogen peroxide at 25^​o​ C is -224 kJ.

2H_​2O_​2(l) -> 2 H_​2O(l) + O_​2 (g)           delta G^​o = -224kJ

a.      Calculate the equilibrium constant for the reaction

b.      What is the chemical significance of the value of the equilibrium constant?

c.       The standard enthalpy change, delta H^​o , for decomposition of hydrogen peroxide is + 196.1 kJ. Determine the standard entropy change, delta S^_​​o, for the reaction at 25^o ​C?

d.      i) what is the significance of the standard free energy change?

         ii) What is the significance of the standard enthalpy change?

         iii) What is the significance of the standard entropy change?

Answer 1

a) 2H_​2O_​2(l) -> 2 H_​2O(l) + O_​2 (g)                 G^​o = -224kJ

G^​o = -RT lnK

-224 x 10³ J = -(8.314 J/K.mol)(298 K) lnK

ln K = 90.41

K = 1.89 x 10³⁹

b) A very high value of K(equilibrium constant) indicates that the reaction goes in the forward direction, to a large extent (almost goes to completion).

c)

-224 kJ = 196.1 kJ -(298 K)(S)

solving the equation for entropy change, we get:

S = 1.41 kJ/K

d. i) negative value for standard free energy change implies that the reacction is spontaneous in forward direction(it is exergonic).

ii) positive value for enthalpy change indicates that the reaction is endothermic.

iii) the positive value for entropy change indicates that there is an increase in entropy during the process. This is evident from the formation of a gaseous product from a liquid reactant. Gases have very high entropy than liquids because in gases, molecules have more movement possible and there is more randomness. Hence, oxygen gas has much high entropy than liquid water and liquid hydrogen peroxide. Thus, entropy increases during the process.

Though, the reaction is endothermic, which doesn't favor spontaneity; but, hight positive entropy change makes the reaction exergonic.