`0.5 mol` of `H_(2)` and `0.5` mol of `I_(2)` react in `10 L` flask at `448^(@)C`. The equilibrium constant `(K_(c ))` is `50` for
`H_(2)+I_(2) hArr 2HI`
a. What is the value of `K_(p)`?
b. Calculate the moles of `I_(2)` at equilibrium.

0.5 mol of H_(2) and 0.5 mole of I_(2) react in 10 litre flast at 448^(@)C . The equilibrium constant K_(c) is 50 for H_(2)(g)+I_(2)(g)hArr2HI(g) a. What is the value of K_(p) b. Calculate mole of I_(2) 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.

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

Finding equilibrium concentrations: A mixture of 0.50 mol H_2 and 0.50 mol I_2 is placed in a 1.00L stainless steel container at 400^@C . The equilibrium constant K_c for the reaction H_2(g)+I_2(g)hArr 2HI(g) is 54.3 at this temperature. Calculate the equilibrium concentrations of H_2 , I_2 , and HI .

A mixture of 0.3 mole of H_(2) and 0.3 mole of I_(2) is allowed to react in a 10 litre evacuated flask at 500^(@) C. The reaction is H_(2) + I_(2) hArr 2H , the K is found to be 64. The amount of undreacted I_(2) at equilibrium is

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