The value of activation energy for the reaction 2HI(g)rarrH_(2)(g)+I_(2)(g) at 581K temperature is 209.5 kJ mol^(-1) ...

The activation energy for the reaction : ltbr. 2Hl(g) rarr H_(2)(g)+I_(2)(g) is 209.5 kJ mol^(-1) at 581K . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?

The activation energy for the reaction, 2Hi(g) to H_(2)(g) + I_(2)(g) is 209.5 kJ moli^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

The equilibrium constant, K for the reaction : 2HI (g) hArr H_(2)(g)+I_(2)(g) at room temperature is 2.85 and that of at 698 K is 1.4 xx 10^(-2) . This implies that the forward reaction is

The equilibrium constant K for the reaction 2HI(g) hArr H_(2)(g)+I_(2)(g) at room temperature is 2.85 and that at 698 K is 1.4 xx10^(-2) . This implies

In the reversible reaction, 2HI(g) hArr H_(2)(g)+I_(2)(g), K_(p) is

For the reaction, 2HI(g) rarr H_(2)(g) + I_(2) (g) - Q KJ , the equilibrium constant depends upon

Calculate the standard free energy change for the reaction: H_(2)(g) +I_(2)(g) rarr 2HI(g), DeltaH^(Theta) = 51.9 kJ mol^(-1) Given: S^(Theta) (H_(2)) = 130.6 J K^(-1) mol^(-1) , S^(Theta) (I_(2)) = 116.7 J K^(-1) mol^(-1) and S^(Theta) (HI) =- 206.8 J K^(-1) mol^(-1) .

Calculate the standard free energy change for the reaction : H_(2)(g) + I_(2)(g) to 2HI(g) DeltaH^(@) = 51.9 kJ "mol"^(-1) Given : S^(@)(H_(2)) = 130.6 J K^(-1) "mol"^(-1) S^(@)(I_(2)) = 116.7 J K^(-1) "mol"^(-1) and S^(@)(HI) = 206.3 J K^(-1) "mol"^(-1) .