This paper contains 7 questions in 5 pages (8 pages with answers). Check the question paper. is believed to follow the following

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Ernest Lyons
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1 DEPARTMENT OF CHEMISTRY, IIT MADRAS CY 101: Equilibrium and Dynamics of Chemical Systems I B. Tech. End Semester Examination; 28 November, 2005 Mars: 60(this part); Time: 3 hours (total) This paper contains 7 questions in 5 pages (8 pages with answers). Chec the question paper. R = (8.314 J = atm dm 3 ) K -1 mol -1 N A = x B =1.38 x J K -1 h = x Js 1. (This question contains 10 short questions totaling 18 mars) I. (2 mars) Data for the reaction, 2 NO + Cl 2 2 NOCl 2, are given: experiment # [NO] (M) [Cl 2 ] (M) initial rate (M s -1 ) (1) x 10-3 (2) x 10-2 (3) x 10-2 Choose the correct option from below: a) zero order in both [NO] and [Cl 2 ] b) 1st order in [NO], 1st order in [Cl 2 ], 2nd order overall c) 1st order in [NO], 2nd order in [Cl 2 ], 3rd order overall d) 2nd order in [NO], 1st order in [Cl 2 ], 3rd order overall e) 2nd order in [NO], 2nd order in [Cl 2 ], 2nd order overall Ans. (c) II. A reaction mechanism: (step1) A + B (step2) C (step3) D + E (step4) F D 3 G C F is believed to follow the following (a) (1 mar) List the intermediate(s) involved in the reaction, if any. Ans: C,D,F (b) (1 mar) Is/are catalyst(s) involved in the reaction? List if any. Ans: None (c) If the steps 1 and 4 are fast and the rate constant 2 << 3, (i) (1 mar) What should be the order of the reaction and with respect to which species? Ans: 1 st order w. r. t. C 1

2 (ii) (1 mar) A plot of which parameter against time should be linear with a negative slope? Ans: ln C III. (1 mar) In Transition State Theory, which one of the following factors is related to the frequency of the activated complex crossing the activation barrier? (a) K / t (b) pz AB (c) r (rate of reaction) (d) B T/h Ans. (d) IV. (2 mars) An opposing reaction of the type R f b P has an equilibrium constant value of 3. Starting at initial concentrations of R at 1 M and P at 0 M, setch the plot of the variation of concentrations of R and P against time, in a time scale such that equilibrium concentrations of both R and P are indicated. Use the following format in your answer boo. [R] 1 M 0.75 M 0.25 M [P] e Concentration [R] e [P] 0 time V. 1 B A 3 2 C D For the above reaction, (i) 2 = 3 1 2

3 (ii) 3 >>> 2 An experiment involving the above reaction was started with the initial concentration of A at 1M. (A) (1 mar) Can steady state approximation be used for the concentration of C? yes (B) (2 mars) What was the concentration of D after infinite time? 0.75 M VI. (1 mar) A chemical reaction taing place in a calorimeter causes the temperature to rise by C. At the end of the reaction, an addition of 50 J of energy to the calorimeter by an electrical heater increases its temperature further by C. What is H of the chemical reaction? Choose the correct option from below. a) 16.7 J b) 16.7 J c) 150 J d) 150 J Ans. (c) VII. (2 mars) Consider the reaction, 2 NO 2 (g) = N 2 O 4 (g) at 27 C, given H = 57.2 J mol -1 and S = Jmol 1 K 1. This reaction is a) Nonspontaneous at 50 0 C, but can be made spontaneous at sufficiently high T. b) Nonspontaneous at 50 0 C, but can be made spontaneous at sufficiently low T. c) Spontaneous at 50 0 C, but can be made nonspontaneous by increasing T. d) Spontaneous at 50 0 C, but can be made nonspontaneous by decreasing T. e) Spontaneous under all temperature conditions. Choose the correct option from above. Ans: (c) VIII. (1 mar) If the entropy of the universe is increasing in any spontaneous process, how can salts crystallize out of solution, a process which causes a greater ordering of the particles? Choose the correct option from below. a) Although the entropy of a system, such as crystallization of dissolved salts in solution, may decrease during a spontaneous process, the entropy of the surroundings increases to a larger extent. b) The crystallization process is a paradoxical violation of the Second Law of Thermodynamics. c) Crystalline salts have higher entropy than their saturated solutions and therefore crystallization is spontaneous. d) Although delta S is negative for crystallisation, G is always negative since H is highly negative. Ans. (a) 3

4 IX. (1 mar) Which of the following options represent the total number of independent variables that are needed to specify a system of C components distributed in P phases at equilibrium? (a) P (C 1) + 2 (b) C (P 1) + 2 (c) C (P 1) (d) C + P 2 Ans. (a) X. (1 mar) Which of the following is NOT a characteristic of a simple two-component eutectic system? (a) Liquid-A and liquid-b are miscible in all proportions. (b) Solid-A and solid-b form a solid solution at eutectic temperature. (c) Intimate mixture of solid-a and solid-b cause mutual lowering of melting points of each other. (d) At the eutectic temperature the system has the lowest degree of freedom. Ans: (b) 2. (7 mars) For the decomposition of N 2 O 5 as per the equation, 2N 2 O 5 (g) 4NO 2 (g) +O 2 (g), the following mechanism is suggested: (step1) N O (step2) NO +NO NO +NO N O (step3) NO+NO NO+O+NO (step4) NO+N O (A) Find the rate law, NO [ ] 2 5 d N O dt consistent with this mechanism. (B) If 3 >> 2, which of the above step(s) contribute(s) to the rate law? 4

5 answer: (A) [ ] d N2O5 = 1[ N2O5] + 2[ NO2][ NO3] 4[ NO][ N2O5] dt d[ NO3 ] = 1[ N2O5] 2[ NO2][ NO3] 3[ NO2][ NO3] = 0 dt d[ NO] = 1[ NO 2][ NO 3] 4[ NO ][ N 2 O 5] = 0 dt d[ N2O5] 213[ N2O5] = dt (5 mars) (B) under the condition, only the first step contributes. (2 mars) 3. (7 mars) Michaelis-Menten Mechanism for an enzyme catalyzed reaction is given as, 1 2 E+S ES P -1 where the notations have their usual significance. Setch the plot of initial rate of the reaction (r 0 ) against initial substrate concentration ([S] 0 ) and indicate (in the plot) the following: (a) The range where the order of the reaction is 0 with respect to [S] 0. (b) The range where the order of the reaction is 1 with respect to [S] 0. (c) The position of (r 0 (max)), where r 0 is maximum. (d) The position of Michaelis-Menten constant ( M ). Answer: (3 mars for the correct plot. 1 mar for each required indication) 5

6 r max r0 (initial rate) (r 0 ) Ini tia r max / 2 l rat e 1 st order wrt [S] 0 0 order wrt [S] 0 M [S] 0 Initial substrate concn 4. (7 mars) Calculate A, U and G for the vaporization of one mole of water at K and one bar pressure. Assume water vapour to be an ideal gas. Heat of vapourisation of water is 40,893 J mol -1. Answer A= U T S G = H T S A= q + w q = w P P P P P G = q q = 0 P P (1mar) RT w P(rev) = P V= P(Vg V) l Vg = P w (rev) = RT = = 3099J A = 3099J (2mars) U = q + w = = 37794J (2mars) (2mars) 5. (7 mars) The gas phase reaction, 2A Decomposition products is bimolecular with an activation energy of 185 J mol 1. A has a molar mass of g and its radius is 1.75 Ǻ. (a) Calculate from collision theory the specific rate constant for the decomposition reaction at 556 K. 6

7 Answer (a) 1/2 πrt 2 AA AA A Z = 2 d N M 1/ = 2 ( ) ( ) = m mol s = dm mol s q = exp = = collision dm m ol s = dm mol s mars (b) If the observed rate constant is 7.25 x 10 8 mol 1 dm 3 s 1, find the steric factor. Answer (b) experimental = dm mol s p = = The value of (col) estimated by students may be slightly different from this. In such a case, their calculation of p can be taen as correct, if the procedure is right. 1 mar (c) If the temperature of the reaction is reduced by a factor of 4, the pre-exponential factor will decrease (increase / decrease) by a factor of 2. (Fill in the blans) 2 mars 6. (7 mars) For the reaction, SnO 2 (s) + 2H 2 (g) Sn (s) + 2H 2 O (g) the equilibrium partial pressure of H 2 was found to be 0.45 atm at 900 K and 0.22 atm at 1100K. The total equilibrium pressure was maintained at 1 atm in a closed vessel. Calculate the following: (A) The standard molar free energy change for the reduction of SnO 2 at 1100 K. At 1100 K, p H2 = 0.22 atm, p H2O = (1 -.22) atm = 0.78 atm K 1100 = (0.78 / 0.22) 2 = G 0 = -RTlnK = J (2 mars) (B) The standard molar enthalpy change for the reduction of SnO 2, assuming it to be constant in the temperature range. 7

8 At 900 K, using a similar procedure, K 900 = 1.49, G 0 = J ln(k 1100 / K 900 ) = ( H 0 /R){( )/(1100x900)} H 0 = J (3 mars) (C) The standard molar entropy change for the reduction of SnO 2 at 900 K. At 900 K, K 900 = 1.49, G 0 = J S 0 = ( H 0 - G 0 ) / T = J K -1 (2 mars) 7. (7 mars) For a chair-chair conformational inversion of cyclohexane, the specific reaction rate is given by, = s / RT 1 e over the temperature range, 117 o C to 24 o C. Calculate the thermodynamic properties, S, H, G at 70 o C. Answer T S E a B R RT = e e e h n T S B e R e n s 1 = at 203 K h BT JK e e h Js 203K s S R = = 34 e = and S = = 2.33 JK mol and E = Jmol a a 1 H = E RT = Jmol 1 (2mars) ( ) G = H T S = = 1 1 (3mars) J mol (2mars) 8

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase changes Apply the second law of thermodynamics to chemical