for the exothermic formation of sulphur trioxide from sulphur dioxide. And oxygen in the gas phase: 2SO_(2)(g) + O_(2)(g)hArr 2SO_(3) (g) K_(p) =40 .5 atm^(-1) at 900 K and Delta H =- 198 kJ (i) Write the expression for the equilibrium constant for the reaction. (ii) At room temperature (~~300 K) will K_(p) be greater than less than or equal to K_(p) at 900 K . (iii) How will the equilibrium be affected if the volume of the vessel contaning the three gases is reduced, keeping the termperature constant ? What happens ? (iv) What is the effect of adding 1 mole of He (g) to a flask containing SO_(2),O_(2) and SO_(3) at equilibrium at constant temperatrue ?

for the exothermic formation of sulphur trioxide from sulphur dioxide.

1.	for the exothermic formation of sulphur trioxide from sulphur dioxide. And oxygen in the gas phase: 2SO_(2)(g) + O_(2)(g)hArr 2SO_(3) (g) K_(p) =40 .5 atm^(-1) at 900 K and Delta H =- 198 kJ (i) Write the expression for the equilibrium constant for the reaction. (ii) At room temperature (~~300 K) will K_(p) be greater than less than or equal to K_(p) at 900 K . (iii) How will the equilibrium be affected if the volume of the vessel contaning the three gases is reduced, keeping the termperature constant ? What happens ? (iv) What is the effect of adding 1 mole of He (g) to a flask containing SO_(2),O_(2) and SO_(3) at equilibrium at constant temperatrue ?
Answer» Solution :(i) The expression for the equilibrium constant `(K_(p))` for the reaction is : `K_(p) =(P_(SO_3)^(2)(G))/(P_(SO_2)^(2)(g)xxP_(O_(2)))=40.5 ATM^(-1)` (ii) the forwardreaction is exothermic and the backward reaction is endothermic in nature. Therefore ,the increase in temperature favour the backward reaction. This means that the `K_(p)` at 300 K will be GREATER than the value at 900K.