For hypothetical reversible reaction 1/2A2(g)+3/2B2(g)⟶AB3(g);ΔH=−20kJ if standard entropies of A2,B2,andAB3are60,40,and50JK−1mol−1, respectively. The above reaction will be equilibrium at
A
400K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
500K
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
250K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
200K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B500K As we know, At equilibrium,ΔStotal=0 ∴S⊖P−S⊖R=0 Also, ΔG=ΔH−TΔS=0 at eqm. so, −20,000=T(50−30−60) T=500K