Q.
The standard free energy change of a reaction is ΔG°=-115 kJ at 298 K. Calculate the value of log10KpR=8.314JK-1 mol-1
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
a
20.16
b
2.303
c
2.016
d
13.83
answer is A.
(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
ΔG°=-RTlnkeq ΔG°=ΔH°-TΔS°=-RTlnkeqlog Kp = 115×10002.303×8.314×298=20.1547≅20.16
Watch 3-min video & get full concept clarity