Answer:
The free energy change for the reaction at 37.0°C is -8.741 kJ.
Explanation:
The free energy of the reaction is given by :
where,
= standard Gibbs free energy
R = Gas constant = 
T = temperature in Kelvins
K = equilibrium constant
We have :
1 kJ = 1000 J
T = 37.0 C = 37 +273.15 K = 310.15 K
Ratio of concentrations of the products to the concentrations of the reactants =K = 21.9
![\Delta G=-16,700 J/mol+8.314J/K mol\times 310.15 K \ln[21.9]](https://tex.z-dn.net/?f=%5CDelta%20G%3D-16%2C700%20J%2Fmol%2B8.314J%2FK%20mol%5Ctimes%20310.15%20K%20%5Cln%5B21.9%5D)
The free energy change for the reaction at 37.0°C is -8.741 kJ.
Answer:
Solution:-
The gas is in the standard temperature and pressure condition i.e. at S.T.P
Therefore,
V
i
=22.4dm
3
V
f
=?
As given that the expansion is isothermal and reversible
∴ΔU=0
Now from first law of thermodynamics,
ΔU=q+w
∵ΔU=0
∴q=–w
Given that the heat is absorbed.
∴q=1000cal
⇒w=−q=−1000cal
Now,
Work done in a reversible isothermal expansion is given by-
w=−nRTln(
V
i
V
f
)
Given:-
T=0℃=273K
n=1 mol
∴1000=−nRTln(
V
i
V
f
)
⇒1000=−1×2.303×2×273×log(
22.4
V
f
)
Explanation:
Answer:
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