For a reversible reaction Kp/Kc =(RT)\(\bigtriangleup\)n at 300K. If Kc/Kp=14941.5 (in litre atm unit) for a given reaction then \(\bigtriangleup\)n for the reaction is  [Given that log Kp/Kc =4.17 and log(24.93)=1.39]

\({Kp \over Kc} = (RT)^{\bigtriangleup n}\)

\(log{Kp \over Kc} = {\bigtriangleup n} \times log(RT)\)

log 14941.5 = Δn × log(0.0821 × 300)

4.17 = Δn × log24.93

4.17 = Δn × 1.39

Δn = 3

---

Kp and Kc Related Questions

1. What is the relation between kp and kc?
2. In the reaction, 2NO ⇌ N2 + O2, are the values of Kc and Kp the same?
3. If, Kp​ = Kc ​× [RT]Δn, for the reaction 2CO2 ​(g) + 2CaCO3 ​(s) ⇌ 2CaC2​(s) + 5O2​(g); Δn is equal to?
4. What is the relation between kp and kc for a gaseous reaction, 2B -> A?

---

Read More: