JEE Main 24 Feb 2021 Shift 1 Solved Paper

For a reaction, A(g)⟶B(g)
Given, Kp (equilibrium constant) =100
‌ Temperature ‌=300K
‌ Pressure ‌=1atm
Formula used, ∆G∘=−RT‌ln‌Kp . . . (i)
Here, ∆G∘= standard Gibb's free energy
R= gas constant =8.31Jmol−1K−1
Put value in Eq (i), we get
∆G∘=−R(300)‌ln‌100
∆G∘=−R(300)(2)‌ln(10)
∵ln(10)=2.3
∆G∘=−R(300)(2)(2.3)
∆G∘=−1380R
‌ Hence, ‌∆G∘=−xR
x=1380