Determine the total number of electrons transferred In this reaction. Consider t

Question

Determine the total number of electrons transferred In this reaction. Consider the first order reaction, C rightarrow products. a) What is the integrated rate law for this reaction? b) If the concentration [C] is measured as a function of time, how can the rate constant for this reaction be determined? c) If the absorbance of light, A, is measured as a function of time, how can the rate constant for this reaction be determined? In this experiment ethanol, CH_3 CH_2 OH, is oxidized by chromic acid, H_2 CrO_4. a) Potassium dichromate, KCr_2 O_7 (aq) reacts with sulfuric acid, H_2 SO_4 (aq) to produce chromic acid H_2 CrO_4 (aq) according to the following reaction. 2 H_3 O+ (aq) + Cr_2 O_7^2- (aq) rightarrow 2 H_2 CrO_4 (aq) + H_2 O(I) The chromic acid H_2 CrO_4 (aq) is made by adding 1.00 mL of 0.0196 M potassium dichromate, KCr_2 O_7 (aq), to 10.0 mL of 3.9 M sulfuric acid, H_2 O_4 (aq). Calculate the concentration of the chromic acid. b) 10.0 uL of CH_3 CH_2 OH is added to 2.00 mL of H_2 CrO_4 (aq). Calculate the concentration of CH_3 CH_2 OH in the resulting solution given that the density of ethanol is 0.789 g/ml.. c) Ethanol is oxidized by the chromic acid according to the following reaction. 6 H_3 O^+ (aq) + 3 CH_3 CH_2 OH(aq) + 2H_2 CrO_4 (aq) rightarrow 3 CH_3 CHO (aq) + 2 Cr^3+ (aq) 14 H_2 O(I) Calculate the concentration of ethanol after reaction goes to completion and the percent

Explanation / Answer

Q4.

V = 1 mL of KCrO7 at M = 0.0196 M

V = 10 mL of H2SO4 at M = 3.9 M

Find concentration of chromic acid produced

mmol of H+= MV = 10*3.9*2 = 78 mmol of H+ present

mmol of Cr2O7- = MV = 1*0.0196 = 0.0196 mmol of Cr2O7-2

cearly, Chromate is limiting

so

ratio is 1 mol of Cr2O7-2 produced 2 mol of H2CrO4, so

0.0196 mol of Cr2O7-2--> 0.0196 *2 = 0.0392 mmol of H2CrO4

Votal at ending = 1+10 = 11

so

[H2CrO4] = mmol/mL = 0.0392 /11 = 0.00356363 M

b)

V = 10 microliters of ethanol ae added to V = 2 mL of acid... find concentation before reaction

C = mass of ethanol / vtotal

mass = D*V = (10*10^-3)(0.789) = 0.00789 grams of ethanol

Vfinal = 2*10^-3 + 10*10^-6 = 0.00201 L 0.00201*10^3 = 2.01 mL

so

C = 0.00789 /2.01

C = 0.00392 grams per mL

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.