Question

D18 Kinetics Review Sheets 8 H*(aq) + 4 Cl(aq) + MnO4 (aq) - 2 Cl(g) + Mn(aq) + 4H2O(l) 1. Cl2 (9) can be generated in the laboratory by reacting potassium permanganate with an acidified solution of sodium chloride. The net-ionic equation for the reaction is given above. a. A 25.00 mL sample of 0.250 M NaCl reacts completely with excess KMnO.(aq). The Cl(a) produced is dried and stored in a sealed container. At 22°C the pressure of the Cl2(g) in the container is 0.950 atm. i. Calculate the number of moles of Cl-(aq) present before any reaction occurs. il. Calculate the volume, in L of the Cl2 (g) in the sealed container. An initial-rate study was performed on the above reaction system. Data for the experiment are given in the table below. Trial [CH] [Mn0,11 [H] Rate of Disappearance of Mno, in MS-' 1 0.0104 0.00400 3.00 2.25 x 10-8 0.0312 | 0.00400 3.00 E EE 2.03 x 10-7 3 0.0312 0.00200 3.00 S E 1.02 x 10-7 Using the information in the table, determine the order of the reaction with respect to each of the following. Justify your answers. I. CI ii. Mno. C. The reaction is known to be third order with respect to H+. Using this information and your answers to part (b) above, complete both of the following: i. Write the rate law for the reaction. il Calculate the value of the rate constant, k, for the reaction, including appropriate units. d. Is it likely that the reaction occurs in a single elementary step? Justify your answer. 2. C Ha(g) reacts readily with HCl(g) to produce C2HCl(9), as represented by the following equation. CH.(g) + HCI(g) → C2H5Cl(g) AH = -72.6 kJ/molan It is proposed that the formation of CzH5Cl(g) proceeds via the following two-step reaction mechanism. Step 1: CH.(g) + HCl(9) CzHs*(g) + CH(g) rate-determining step (thus slow) Step 2: CzHs*(g) + CH(g) → CzHsCl(g) fast step a. Write the rate law for the reaction that is consistent with the reaction mechanism above. b. Identify an intermediate in the reaction mechanism above.

Answer 1

Q1(a)

(i) the number of moles of Cl^- present before the reaction = 0.250mol/L ×25.00×10^-3L = 6.25×10^-3 mol. (Answer)

(ii)

Number of moles of chlorine gas formed = (2/4)×moles of Cl^- = 6.25/2 × 10^-3 mol

V = nRT/P = 3.125×10^-3mol × 0.0821atm-L/K.mol × 295.15K/0.950atm = 0.0797 L

(b)

(i) order of reaction w.r.t. Cl^- = 2

As the concentration of chloride ions is increased thrice , the rate of reaction increases nine times (i.e. 3² ) .

(ii) order of reaction w.r.t. MnO₄^- = 1

As the concentration of MnO₄^- is decreased to half , the rate of reaction deceased to its half (i.e. (1/2)¹) .

(c)

(i) rate of the reaction = k.[H⁺]³[Cl^-]²[MnO₄^-]¹

(ii) k = 2.25×10^-8 M.s^-1/(3.00 M)³.(0.0104M)² × 0.00400M)

= 0.001926 M^-5 .s^-1

= 1.93×10^-3 M^-5.s^-1

(d)

The overall oder of reaction is 3 +2 +1 = 6 ( H⁺ acts as catalyst ) . Order w.r.t. reactants = 2+1 = 3

The order of reaction is higher , so the reaction is not likely to be occurring in single step or the reaction is less likely to be elementary reaction .

Q2:

(a)

Rate law is determined by the slow step in reaction mechanism .

Rate of reaction = k.[C₂H₄][HCl]

(b)

The intermediate in the given mechanism is C₂H₅⁺ ( Answer)