In an experiment, 0.01 mole of CH3CO2H(l), 0.02 mole of C2H5OH(l) and 0.005 mole of H2SO4 are heated. The temperature of the mixture is kept at 300 C for 1 hour. After cooling, the reaction mixture is titrated with 0.50M NaOH solution, using a suitable indicator. It requires 35.2 cm3 NaOH solution to reach the end-point.
Questions:
1. Write an equation to show the reaction of CH3CO2H(l) and C2H5OH(l).
(given) CH3CO2H(l) + C2H5OH(l) <=> CH3CO2H5H(l) + H2O(l)
2. What is the function of concentrated H2SO4 in the reaction?
3. What is the function of cooling?
4.Calculate the number of mole of the following substance at eqm at 300 C.
a) CH3CO2H(l)
b)C2H5OH(l)
c)CH3CO2H5H(l)
d)H2O(l)
5. Determine the equilibrium constant, Kc, of the reaction at 300 C.
Chemical Equilibrium
2011-05-08 11:29 pm
回答 (2)
2011-05-09 7:25 am
✔ 最佳答案
1.CH3CO2H(l) + C2H5OH(l) ⇌ CH3CO2C2H5(l) + H2O(l)
2.
It acts as a catalyst of the reaction.
3.
The reaction mixture is cooled down in order to make the reaction practically stop. Hence, it can make sure that the concentration of each component (the equilibrium position) remains the same as that at 300°C
4.
a)
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
CH3CO2H(aq) + NaOH(aq) → CH3CO2Na(aq) + H2O(l)
No. of moles of NaOH used = 0.50 * (35.2/1000) = 0.0176 mol
No. of moles of H2SO4 = 0.005 mol (H2SO4 as a catalyst)
No. of moles of NaOH reacted with H2SO4 = 0.005 * 2 = 0.01 mol
No. of moles of NaOH reacted with CH3CO2H = 0.0176 - 0.01 = 0.0076 mol
No. moles of CH3CO2H at eqm = 0.0076 mol
b)
No. of moles of CH3CO2H reacted = 0.01 - 0.0076 = 0.0024 mol
No. of moles of CH3CH2OH at eqm = 0.02 - 0.0024 = 0.0176 mol
c)
No. of moles of CH3CO2C2H5 at eqm = 0.0024 mol
d)
No. of moles of H2O at eqm = 0.0024 mol
5.
Equilibrium constant, Kc = (0.0076*0.0176) / (0.0024)² = 23.2
2011-05-14 12:45 am
有兩個問題，望各位指教：
一，冷凍時，平衡位置不變?
二，NaOH為強鹼，而濃度不算低，不能水解乙酸乙酯?
一，冷凍時，平衡位置不變?
二，NaOH為強鹼，而濃度不算低，不能水解乙酸乙酯?
收錄日期: 2021-04-23 22:05:24
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