The answer is B...
because any of the others could have an opinion based answer but B will always be the same its a fact not opinion.
To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40
Answer: 26.5 mm Hg
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 
= ?
= final pressure at 
= 100 mm Hg
= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature =
Now put all the given values in this formula, we get
![\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BP_1%7D%7B100%7D%29%3D%5Cfrac%7B28000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B299.5%7D-%5Cfrac%7B1%7D%7B267.9%7D%5D)
Thus the vapor pressure of 
in mmHg at 26.5 ∘C is 26.5
