Question

7.43 hours b) 16.5 hours c) 20.7 hours d) 34.2 hours e) 45.3 hours P21. (See. 13.4) A certain reaction has artes at 205 lata temperature of 25 .11 the activation energy (E) equals 85.6 kJ/mol what is the frequency factor (A) for the reaction (R = 8.314 J/mol K) a) 6.022 x 1023 /s b) 39.3/s 2.073 x 1017/s d) 7.453 x 106/s e) 5.128 x 1011 /s P22. (Sec. 13.4) The following reaction is a first order reaction. Given the following two rate constants and temperatures, calculate the activation energy (E). in kJ/mol, for the decomposition of nitrous oxide, N20. [N2O(g) N2(g) + O2(g)] (At 35 °C, Rate Constant: 0.00014 s) & (At 45 °C, Rate Constant: 0.0005 s) (R-8.314 J/mol - K) a) 81.0 kJ/mol b) 97.1 kJ/mol c) 103.6 kJ/mol d) 211.2 kJ/mol e) 303.0 kJ/mol P23. (Sec. 13.4) The following reaction is a second order reaction. Given the following two ra constants and temperatures, calculate the activation energy (Ea), in kJ/mol, for the decompositi of acetaldehyde. (C2H.O(g) → CH.(g) + CO(g)] (At 486 °C, Rate Constant: 0.00105 M".s) & (At 563 °C, Rate Constant: 0.0214 M .s' (R = 8.314 J/mol. K) a) 8.92 kJ/mol b) 411 kJ/mol c) 163.6 kJ/mol d) 87.2 kJ/mol e) 206.5 kJ/mol

Answer 1

Arhenius Theory K= A e-Ea/RT & Tele A = frequency factor - SEK = 205 stes T = 25°c = 298k E = 85.6 kJ/mol I RE 8.314 7 molok as we can simplify equation - lk = ln A - Ea In 205 = ln A = obsob 8.314 X 298 x 103 5 mins

5.32 = A - 34.55 In A = 39.87 A = 2.0668 1017 correct option is ☺ so the RLT Tz T = 35°c T₂ = 45°6 Athenius equation = 0.0ool45-1 k, = 0.00058 il R = 8.314 J mol.k. ln - 10.0005 0.00014 Ea 1 8.314 x 103 | 308 - 1 318. 1.27 = Ear 8.314x103 318 - 308 7 308x318 1:27 = Eart 10 8.314x 10's & 97944 814 10 187 = EXO & = 1.27X 814 Ea = 10313 kJ/mol so the correct option is (C)

T = 466.c. Tz = 563°c P23 L k, = 0.00105 mist K = 0.0214 m-'s E In (0.0aly) = 10.0214 10.00105 x (159 159 8.314810 3 Ea 8.314 x 10-s 77 634524 77 3 - Ea x 17 5275.43 1 Ea = 345275.43 Ea = 205.53 kJ moli so the correct option is that @