For the reaction `2HI hArr H_(2)+I_(2)`

For the reaction 2 HI hArr H_(2) + I_(2) the equilibrium constant K at 440^(@)C is 0.022 . The equilibrium constant for I_(2) + H_(2) hArr 2 HI is

If the equilibrium constant of the reaction 2HI hArrH_(2) + I_(2) is 0.25 , then the equilibrium constant of the reaction H_(2) + I_(2)hArr2HI would be

For a reaction 2HI hArr H_(2)+I_(2) , at equilibrium 7.8 g, 203.2 g , and 1638.4 g of H_(2), I_(2) , and HI, respectively were found. Calculate K_(c) .

For a reaction 2HI hArr H_(2)+I_(2) , at equilibrium 7.8 g, 203.2 g , and 1638.4 g of H_(2), I_(2) , and HI, respectively were found. Calculate K_(c) .

If the equilibrium constant of the reaction 2HIhArr H_(2)+I_(2) is 0.25, then the equilibrium constant for the reaction, H_(2)(g)+I_(2)(g)hArr 2HI(g) would be

If the equilibrium constant of the reaction 2HIhArr H_(2)+I_(2) is 0.25, then the equilibrium constant for the reaction, H_(2)(g)+I_(2)(g)hArr 2HI(g) would be

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

If DeltaG^(@)[HI(g)=-1.7kJ] , the equilibrium constant for the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g) at 25^(@)C 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