The vapour pressure of water at 25°C is 0.0313 atm. Calculate the values of `K_(p)` and Kc at 25°C for the equilibrium `H_(2)O(1)hArr H_(2)O(g)`.

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

Equilibrium constants are given (in atm) for the following reactions at 0^@C (a) SrCl_2.6H_2O(s) iff SrCl_2.2H_2O(s) + 4H_2.O(g), K_p = 6.89 xx 10^(-12) (b) Na_2HPO_4.12H_2O(s) iff Na_2HPO_4. 7H_2O(s) +5H_2O(g), K_p = 5.25 xx 10^(-13) (c) Na_2SO_4. 10 H_2O(s) iff Na_2SO_4(s) + 10H_2O(g), K_p = 4.08 xx 10^(-25) The vapour pressure of water at 0^@C is 4.58 torr. Calculate the pressure of water vapour in equilibrium at 0^@C with each of (a), (b) and (c)

A mixture of 4 moles H_(2)O and 1 mole CO is allowed to react to come to an equilibrium. The equilibrium pressure is 2.0 atm. Calculate partial pressure (in atm) of CO(g) at equilibrium. H_(2)O (g) + CO(g) harr CO_(2) (g) + H_(2) (g); K_(C) =0.5 Use sqrt(41) = 6.4

Calculate the value of K_c for the given reactions: N_2O_4(g)hArr 2NO_2(g),K_p=11 ( at 100^(@)C )

The vapour pressure of water at 25^(@)C is 18 mmHg. If 18 g of glucose (C_(6)H_(12)O_(6)) is added to 178.2 g of water at 20^(@)C , the vapour pressure of the resulting solution will be :

The vapour pressure of water at 25^(@)C is 18 mmHg. If 18 g of glucose (C_(6)H_(12)O_(6)) is added to 178.2 g of water at 20^(@)C , the vapour pressure of the resulting solution will be :

K_p/K_c for the reaction: CO(g)+1/2O_2(g) hArr CO_2(g) is :