Let assume initial moles is 1
A2→2A
t=0 1
t=t 1−0.20.2×2
PartialpressureofA2=1.20.8×1
PartialpressureofA=1.20.4×1
kp=(1.20.8)×1(1.20.4)2×1=61
ΔG∘=−RTlnk=−8.314×320ln61
=8.314×320(0.693+1.098)
⋍4763J/mole
At 320 K, a gas A2 is 20 % dissociated to A(g) . The standard Gibbs free energy change at 320Kand1atminJmol–1 is approximately: (R=8.314JK–1mol–1;ln2=0.693;ln3=1.098)
Held on 16 Apr 2018 · Verified 6 Jul 2026.
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