​ Kc for the reaction SO2 + 1/2 O2 ⇌ SO3 at 600º C is 61.7. Calculate Kp. What is the unit of Kp for the above equilibrium? ​

Question

K_c for the reaction SO₂ + \(\frac{1}{2}\) O₂ ⇌ SO₃ at 600º C is 61.7. Calculate K_p. What is the unit of K_p for the above equilibrium? (R = 0.0821 L atm degree^-1 mol^-1 )

Answer 1

For the reaction 

SO₂ + \(\frac{1}{2}\) O₂ ⇌ SO₃ 

Given, K_c = 61.7 

T = 600 + 273 = 873K 

∆n = No. of moles of product - No. of moles of Reactants 

= 1 − (1 + \(\frac{1}{2}\) ) 

= −  \(\frac{1}{2}\)

∴ K_p = K_c (RT)^∆ 

= 61.7(0.0821 × 873)^{−\(\frac{1}{2}\)} 

= 7.29 

K_p = \(\frac{(P_{SO_3})}{(P_{SO_2}).(P_{O_2})^{\frac{1}{2}}}\)

= \(\frac{atm}{atm.(atm)^{\frac{1}{2}}}\)

= \(atm^{-\frac{1}{2}}\) 

∴ The unit of Kp for the above equilibrium is \(atm^{-\frac{1}{2}}\)