The variation of equilibrium constant with temperature is given below Temperature Equilibrium Constant T1=25∘CK1=10T2=100∘CK2=100 The values of ΔH0,ΔG0 at T1 and ΔG0 at T2 in kJmol−1 respectively, are close to [use R=8.314JK−1mol−1

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

0.64,−5.71 and −14.29

0.64,−7.14 and −5.71

28.4,−5.71 and −14.29

28.4,−7.14 and −5.71

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

log10⁡K2K1=ΔH02.303R1T1−1T2⇒log⁡(10)=ΔH02.303R1298−1373⇒ΔH02.303×8.314=298×37375⇒ΔH0=28377.1J/mol=28.4 kJmolat T1ΔG0=−RT1×2.303log10⁡K1=−2.303×8.314×298log1010=−5.71kJ/mol At T2ΔG0=−2.303RT2log⁡K2=−2.303×8.314×373×2=−14.29 kJmol

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET