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Question

For a reversible reaction KpKc=(RT)Δn at 300K. If KpKc=14941.5 (in litre atm unit) for a given reaction then Δn for the reaction is :
[Given that logKpKc=4.17 and log(24.93)=1.39]

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Solution

KpKc=(RT)Δn
logKpKc=Δn×log(RT)
log14941.5=Δn×log(0.0821×300)
4.17=Δn×log24.93
4.17=Δn×1.39
Δn=3

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