Answer : The value of 
for this reaction is, 
Explanation :
The given chemical reaction is:
Now we have to calculate value of 
.
![\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5Bn_%7BHCH_3CO_2%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28HCH_3CO_2%28g%29%29%7D%5D-%5Bn_%7BCH_3OH%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CH_3OH%28g%29%29%7D%2Bn_%7BCO%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CO%28g%29%29%7D%5D)
where,
= Gibbs free energy of reaction = ?
n = number of moles
= -389.8 kJ/mol
= -161.96 kJ/mol
= -137.2 kJ/mol
Now put all the given values in this expression, we get:
![\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5B1mole%5Ctimes%20%28-389.8kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%28-163.2kJ%2Fmol%29%2B1mole%5Ctimes%20%28-137.2kJ%2Fmol%29%5D)
The relation between the equilibrium constant and standard Gibbs, free energy is:
where,
= standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol
R = gas constant = 8.314 J/L.atm
T = temperature = 
= equilibrium constant = ?
Now put all the given values in this expression, we get:
Thus, the value of for this reaction is,
Please be specific. Which umbrella?
Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
Answer:
x-10 should be the answer
Answer: The answer is D. This has a Carboxylic Acid group, and is acetic acid, or Ethanoic Acid.
ALWAYS LOOK for the Functional Group in question.
A. Would likely not stay in water, or at least not be acidic, for it is butane gas.
B. Is 1-propanol, and alcohols are not acidic as a rule. Certainly not in water.
C. This is an Ether. It will not give up an H+, it it not an acid.
E. This functional group is an amine, which is more “base” like, since the lone pairs of the Nitrogen atom would tend to attract a H+.
