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

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For the reaction 

SO2 + \(\frac{1}{2}\) O2 ⇌ SO3 

Given, Kc = 61.7 

T = 600 + 273 = 873K 

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

= 1 − (1 + \(\frac{1}{2}\) ) 

= −  \(\frac{1}{2}\)

∴ Kp = Kc (RT) 

= 61.7(0.0821 × 873)\(\frac{1}{2}\) 

= 7.29 

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

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

\(atm^{-\frac{1}{2}}\) 

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

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