chapter13_part1

# chapter13_part1 - 560 Chapter 13 Chemical Kinetics Rates of...

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Chapter 13: Chemical Kinetics: Rates of Reactions 560 End-of-Chapter Solutions for Chapter 13 Summary Problem (a) Answer: –101.3 kJ/mol Strategy and Explanation: Given the values of the rate constant and several temperatures, determine the activation energy, the frequency factor and the rate constant at a new temperature. As described in Section 13.5 page 633, we can make a linear graph like Figure 13.11 by taking the logarithm of the rate constant and the reciprocal of the absolute temperature. The slope is related to the activation energy and the y- intercept is related to the frequency factor. The Arrhenius equation can then be used to find the rate constant at a new temperature. Convert temperatures to Kelvin: °C + 273.15 = Kelvin. T (°C) T (K) 1/T (K –1 ) k (s –1 ) ln(k) 47.3 320.5 0.003121 0.233 – 1.457 50.9 324.1 0.003086 0.354 – 1.038 55.0 328.2 0.003047 0.606 – 0.501 60.0 333.2 0.003002 1.022 0.02176 66.0 339.2 0.002949 1.873 0.6275 The slope of this graph is related to E a according to this equation: m = – E a R . with R = 0.008314 kJ/mol K E a = –mR = –(–12190 K –1 )(0.008314kJ/mol K) = 28.6 kJ/mol Reasonable Answer Check: The linearity of the graph is satisfying. Because the slope is negative, and the equation indicates that the slope has a negative relationship with E a , it makes sense that the reaction is endothermic. (b) Answer: 0.364 g Ni(CO) 4 Strategy and Explanation: Given a balanced chemical equation and the masses of both reactants, determine the maximum quantity of the product produced.

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Chapter 13: Chemical Kinetics: Rates of Reactions 561 The production of Ni(CO) 4 from Ni and CO is the reverse of the equation given in the question. Ni(s) + 4 CO(g) Ni(CO) 4 (g) Calculate moles and identify the limiting reactant as described in Section 4.5, then use the moles of product produced from the limiting reactant and the molar mass of the product to find the grams. The balanced equation says: 4 mol CO produces 1 mol Ni(CO) 4 . 2.05 g CO " 1 mol CO 28.0101 g CO " 1 mol Ni(CO) 4 4 mol CO = 0.0183 mol Ni(CO) 4 The balanced equation says: 1 mol Ni produces 1 mol Ni(CO) 4 . 0.125 g Ni " 1 mol Ni 58.6934 g Ni " 1 mol Ni(CO) 4 1 mol Ni = 0.00312 mol Ni(CO) 4 The number of Ni(CO) 4 moles produced from Ni is smaller (0.00213 mol < 0.0183 mol), so Ni is the limiting reactant. Find the mass of 0.00213 mol Ni(CO) 4 : 0.00213 mol Ni(CO) 4 " 170.7338 g Ni(CO) 4 1 mol Ni(CO) 4 = 0.364 g Ni(CO) 4 (c) Answer: first step: unimolecular, second step: bimolecular Strategy and Explanation: A reaction with exactly one reactant is unimolecular and elementary. A reaction with exactly two reactants, either two of the same reactants or one of each of two different reactants, is bimolecular and elementary. A reaction that has more than two reactants, in any combination, is not elementary. The first step has one reactant (the Ni(CO)
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chapter13_part1 - 560 Chapter 13 Chemical Kinetics Rates of...

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