<u>Answer:</u> The concentration of ethyl ethanoate at equilibrium is 
<u>Explanation:</u>
Given values:
Equilibrium concentration of ethanol = 
Equilibrium concentration of ethanoic acid =
The given chemical equation follows:
The expression of 
for above equation follows:
![K_c=\frac{[C_2H_5OH][CH_3COOH]}{[CH_3COOC_2H_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC_2H_5OH%5D%5BCH_3COOH%5D%7D%7B%5BCH_3COOC_2H_5%5D%7D)
Putting values in above expression, we get:
![0.27=\frac{0.42\times 0.42}{[CH_3COOC_2H_5]}](https://tex.z-dn.net/?f=0.27%3D%5Cfrac%7B0.42%5Ctimes%200.42%7D%7B%5BCH_3COOC_2H_5%5D%7D)
![[CH_3COOC_2H_5]=\frac{0.42\times 0.42}{0.27}=0.653mol/dm^3](https://tex.z-dn.net/?f=%5BCH_3COOC_2H_5%5D%3D%5Cfrac%7B0.42%5Ctimes%200.42%7D%7B0.27%7D%3D0.653mol%2Fdm%5E3)
Hence, the concentration of ethyl ethanoate at equilibrium is
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
Answer:
-252.5 kJ/mol = ΔH H2O(g)
Explanation:
ΔH Fe2O3 = -825.5kJ/mol
ΔH H2 = 0kJ/mol
ΔH Fe = 0kJ/mol
Based on Hess's law, ΔH of a reaction is the sum of ΔH of products - ΔH of reactants. For the reaction:
Fe2O3(s) + 3 H2(g) →2Fe(s) + 3 H2O(g)
ΔHr = 67.9kJ/mol = 3*ΔH H2O + 2*ΔHFe - (ΔH Fe2O3 + 3*Δ H2)
67.9kJ/mol = 3*ΔH H2O + 2*0kJ/mol - (ΔH -825.5kJ/mol + 3*Δ H2)
67.9 = 3*ΔH H2O(g) + 825.5kJ/mol
-757.6kJ/mol = 3*ΔH H2O(g)
<h3>-252.5 kJ/mol = ΔH H2O(g)</h3>
