exhibits the rate law where = 1.0 10–5 s–1 at 25°C. This reaction is run where the initial concentration of NOBr k  M–1 ([NOBr]0) is 0.39 What is one half-life for this experiment? M. a) 3.9 s b) 2.0 10–5 s  c) 6.9  10–4 s d) 2.6 105 s  e) none of these 70. The reaction 2NOBr 2NO + Br2  exhibits the rate law Rate = = k[NOBr]2 – where = 1.0  10–5  s–1 at 25°C. This reaction is run where the initial concentration of NOBr k M–1 ([NOBr]0) is 1.00 10–1  M. The [NO] 1.00 hour has passed is a) 3.6 10–4  M b) 9.9 10–3  M c) 9.7 10–3  M d) 1.0  10–3 M e) none of these Use the following to answer questions 71-72: For the reaction A Products, successive half-lives are observed to be 10.0 min and 40.0 min.  71. The reaction follows the integrated rate law a) [A] = + [A]0 –kt b) ln [A] = + ln [A]0 –kt c) = + kt d) = + kt e) none of these 72. At the beginning of the reaction, [A] was 0.35 The numerical of the rate constant is M. a) 0.069 b) 0.29 c) 10. d) 0.018 e) none of these 73. The reaction is first order in N2O5. For this reaction at 45oC, the rate constant = 1.0 10–5 s–1, where the rate law is  k defined as For a particular experiment ([N2O5]0 = 1.0 10–3 calculate [N2O5] after 3.8 105 seconds.  M), 
exhibits the rate law where k = 1.0  10–5 M–1 s–1 at 25°C. This reaction is run where the initial concentration of NOBr ([NOBr]0) is 0.39 M. What is one half-life for this experiment? a) 3.9 s b) 2.0  10–5 s c) 6.9  10–4 s d) 2.6  105 s e) none of these 70. The reaction 2NOBr  2NO + Br2 exhibits the rate law Rate = k[NOBr]2 = – where k = 1.0  10–5 M–1  s–1 at 25°C. This reaction is run where the initial concentration of NOBr ([NOBr]0) is 1.00  10–1 M. The [NO] 1.00 hour has passed is a) 3.6  10–4 M b) 9.9  10–3 M c) 9.7  10–3 M d) 1.0  10–3 M e) none of these Use the following to answer questions 71-72: For the reaction A  Products, successive half-lives are observed to be 10.0 min and 40.0 min. 71. The reaction follows the integrated rate law a) [A] = –kt + [A]0 b) ln [A] = –kt + ln [A]0 c) = kt + d) = kt + e) none of these 72. At the beginning of the reaction, [A] was 0.35 M. The numerical of the rate constant is a) 0.069 b) 0.29 c) 10. d) 0.018 e) none of these 73. The reaction is first order in N2O5. For this reaction at 45oC, the rate constant k = 1.0  10–5 s–1, where the rate law is defined as For a particular experiment ([N2O5]0 = 1.0  10–3 M), calculate [N2O5] after 3.8  105 seconds.