Question

Q.31 For exothermic formation of sulphur trioxide from sulphur dioxide and oxygen in gas phase:
2502 (g) + O2(g) → 2503 (g)
Kp = 40.5 /atm AT 900 K and ∆H=-198 kJ
a. Write the expression for equilibrium constant for the reaction.
b. At room temperature ( = 300 K) will Ko be greater than , less than or equal to K, at 900 K.
c. How will the equilibrium affected if the volume of the vessel containing three gases is reduced,
keeping the temperature constant? What happens?
d. What is the effect on adding 1 mole of He(g) to the flask containing SO2, O2, SOz at equilibrium at
constant temperature.
e. What is common ion effect?​

Answer 1

Answer:

Solution :

(i) The expression for the equilibrium constant

(Kp)

for the reaction is : <br>

Kp= P2so3 = 40.5atm -1

P2so2(g) x Po2

<br> (ii) the forward reaction is exothermic and the backward reaction is endothermic in nature. Therefore ,the increase in temperature favour the backward reaction. This means that the

at 300 K will be greater than the value at 900K.