JEE Main 2022 Chemistry question
The standard free energy change for 50% dissociation of into at 27°C and 1 atm pressure is −x J . The value of x is . (Nearest Integer) [Given : R = 8.31 J , log 1.33 = 0.1239, ln 10 = 2.3]
Answer: 710
Solution
Starting from 1 mol of with 50% dissociation: 0.5 mol and 1 mol , i.e. total 1.5 mol, so at 1 atm the partial pressures are 0.33 atm for
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