Hypo Chemistry !!
JSR
In a hypothetical world of CHEMISTRY, The Standard free energy is defined at 0 Kelvin & 1 atm pressure.
While a scientist was doing experiment (at a pressure of 1 atm and with all ideal conditions ) In his reaction , the Enthalpy varied as T raised to power 3 .
The spontaneity (S) varied as T.
Then find the value of Equilibrium constant (k) When reaction goes from 0 Kelvin to 6 Kelvin respectively.
Answer would be in format , e raised to power x (where x is integer )
Here T defines Temperature And All the quantities are in their respective standard units.
Y = |x| + 4 divided by 100. ( | | represents modulus function )
Then type Y as your answer.
Take value of R (universal constant ) be 1 divided by 12 (1/12) Also , if you think the conditions or question has some flaw , you may enter your answer as 0.
Thanks for having look at my question. ( sorry for inconvenience as I don’t know to use much Latex)
The answer is 4.
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1 solution
Using the equation, dG=dH-TdS Integrating both side w.r.t to T (Temperature) We get G - Gs = T ^3 - (T^2/2) Gs represent Standard Free Energy. Using limits from 0 Kelvin to 6 Kelvin we get G- Gs = 196
Using formula, G = Gs -RTlnK Substituting values, We get k = e raised to power negative of 396
|x|=396 Y =396 +4 /100 Y= 400/100 Y=4