Question

ANSWER ALL QUESTIONS QUESTION 1 [20 MARKS] Equilibrium constant, Ke for ethanoic acid esterification (CH3COOH) with ethanol (C2H5OH) is 4.0 at room temperature. Calculate the composition of the equilibrium mixture if 100 g ethanoic acid, 100 g ethanol and 36 g of water are allowed to achieve equilibrium. (Pemalar keseimbangan, K bagi pengesteran asid etanoik (CH COOH) dengan etanol (CH5OH) ialah 4.0 pada suhu bilik. Hitung komposisi campuran keseimbangan jika 100 g asid etanoik, 100 g etanol dan 36 g air dibiarkan mencapai keseimbangan) (12 marks) b) The hydrolysis experiment of methyl propanoate (CH3CH2COOCH3) to produce propanoic acid (CH3CH2COOH) and methanol (CH3OH) was performed at room temperature under acidic condition. Four solutions (Table 1) with 10.00 mL aliquot of each solution were prepared and titrated with 1.00 M natrium hydroxide (NaOH). (Eksperimen hidrolisis metilpropanoat (CHOCH COOCH3) untuk menghasilkan asid propanoik (CH3CH2COOH) dan metanol (CH3OH) telah dilakukan pada suhu bilik di dalam keadaan berasid. Empat larutan (Jadual 1) dengan 10.00 ml alikuot bagi setiap larutan telah disediakan dan dititratkan dengan 1.00 M natrium hidroksida (NaOH)) Solution HCI 5.00 5.00 5.00 H20 5.00 0.00 0.00 0.00 Table 1 Volume (ml) CH3CH2COOCH3 CH3OH 0.00 0.00 5.00 0.00 4.00 1.00 0.00 CH3CH2COOH 0.00 0.00 0.00 5.00 4.00 1.00 Given: Substance На H20 CH3CH2COOCH Density (g mL) 1.19 1.00 0.92 0.79 0.99 Molar mass (g mol-) 36.46 18.02 88.11 32.04 74.08 CH3OH CH3CH2COOH KIM3102 SULIT CONFIDENTIAL Determine the number of mole (n) of NaOH to react with HCl from Solution A when 16.5 mL of NaOH is required to achieve the equilibrium of this solution. (Tentukan bilangan mol (n) NaOH untuk bertindakbalas dengan HCl dari Larutan A apabila 16.5 ml NaOH diperlukan bagi larutan ini mencapai keseimbangan.) (3 marks) Calculate the equilibrium constant (Kea) of the reaction in Solution B; if Solution B requires 9.20 mL of natrium hydroxide (NaOH). Concentration of water is assumed to be constant throughout the reaction. (Kirakan pemalar keseimbangan (Keo) untuk tindakbalas di dalam Larutan B; sekiranya Larutan B memerlukan 9.20 ml natrium hidroksida (NaOH) bagi

Answer 1

1.(a). Sol :-

Number of moles = Given mass in g / Gram molar mass

So,

Number of moles of ethanoic acid = 100 g / 60.052 g/mol

= 1.665 mol

Number of moles of ethanol = 100 g / 46.07 g/mol

= 2.171 mol

and

Number of moles of water = 36 g / 18 g/mol

= 2.0 mol

The reaction and ICE table is :

.........................CH₃COOH (aq) .........+............CH₃CH₂OH (aq) <----------> CH₃COOCH₂CH₃ (aq).........+.......H₂O (l)

Initial (I)..............1.665 mol.....................................2.171 mol............................0.0 mol.........................................2.0 mol

Change (C)...........-α................................................-α........................................+α..................................................+α

Equilibrium (E).......(1.665-α) mol..........................(2.171-α) mol............................α mol.....................................(2.0+α) mol

Here, α = Amount dissociated per mole

Expression of concentration equilibrium constant (K_c) is :

K_c = [CH₃COOCH₂CH₃].[H₂O] / [CH₃COOH].[CH₃CH₂OH]

4.0 = α(2.0+α) / (1.665-α)(2.171-α)

4.0(1.665-α)((2.171-α) = α(2.0+α)

(6.66 - 4.0 α)(2.171-α) = 2.0 α + α²

14.45886 - 15.344 α + 4.0 α² = 2.0 α + α²

3.0 α² - 17.344 α + 14.45886 = 0

On solving this equation :

α = 1.01015

Now, The equilibrium moles of all the species are :

Moles of ethanoic acid = (1.665-1.01015) mol = 0.65485 mol

Moles of ethanol = (2.171-1.01015) mol = 1.6085 mol

Moles of ethylethanoate = α = 1.01015 mol

and

Moles of water = (2.0 + 1.01015) mol = 3.01015 mol

Also,

The equilibrium mass of all the species are :

Mass of ethanoic acid = 0.65485 mol x 60.052 g/mol = 39.325 g Mass of ethanol = 1.6085 mol x 46.07 g/mol = 74.104 g Mass of ethylethanoate = 1.01015 mol x 88.11 g/mol = 89.00 g and Mass of water = 3.01015 mol x 18 g/mol = 54.183 g