`H_(2)(g) + I_(2)(g)hArr2HI(g)` When 46 g of `I_(2)` and 1 g of `H_(2)` gas heated at equilibrium at `450^(@)C`, the equilibrium mixture contained `1.9` g of `I_(2)` . How many moles of `I_(2)` and HI are present at equilibrium ?

0.5 mol of H_(2) and 0.5 mole of I_(2) react in 10 litre flask at 448^(@)C . The equilibrium brium constant K_(C) is 50 for H_(2)(g)I_(2)(g) hArr 2HI(g) Calcualte mole of I_(2) at equilibrium.

At 700K equilibrium constant for the reaction : H_(2)(g) +I_(2)(g)hArr2HI(g) is 54.8 . If 0.5mol^(-1) of HI(g) is present at equilibrium at 700K . What are the concentration of H_(2)(g) and I_(2)(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K ?

For the equilibrium reaction H_(2(g))+I_(2(g))hArr 2HI_((g))

For the equilibrium reaction H_(2(g))+I_(2(g))hArr 2HI_((g))

For the equilibrium reaction H_(2(g))+I_(2(g))hArr 2HI_((g))

For the equilibrium reaction H_(2(g))+I_(2(g))hArr 2HI_((g))

At 500 K, the equilibrium costant for the reaction H_(2(g))+I_(2(g))hArr2HI_((g))" is "24.8" If "(1)/(2)mol//L of HI is present at equilibrium, what are the concentrations of H_(2)andI_(2) , assuming that we started by taking HI and reached the equilibrium at 500 K ?

At 500 K, the equilibrium costant for the reaction H_(2(g))+I_(2(g))hArr2HI_((g))" is "24.8" If "(1)/(2)mol//L of HI is present at equilibrium, what are the concentrations of H_(2)andI_(2) , assuming that we started by taking HI and reached the equilibrium at 500 K ?

The value of K_(c) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) is 64 at 773K . If one "mole" of H_(2) , one mole of I_(2) , and three moles of HI are taken in a 1 L flask, find the concentrations of I_(2) and HI at equilibrium at 773 K .