CHEM 110 Final Exam Chemistry (2024/2025) Questions with Verified Answers.

CHEM 110 Final Exam Chemistry (2024/2025) Questions with Verified Answers. Part 1. The questions in this part to be answered on the scan sheet provided. Each question is worth one mark. Questions 1 and 2 refer to the following information: The concentration of gaseous nitrogen dioxide (NO2(g)) was measured as NO2(g) decomposes to form NO(g) and O2(g) according to the reaction: NO2(g) NO(g) + O2(g) The following plot of 1/[NO2] versus time is shown below and the line of best fit yielded a straight line through all the data points. 1. From these data, you can deduce that this reaction is (c) second order with respect to NO2(g) The y-axis is plotted with 1/[NO2] => 2nd order If [H2O2] is plotted on y-axis then the reaction is zero order If 1/[ H2O2] is plotted on y-axis then the reaction is second order 2. Given the following information regarding points A and B on the graph 0 10 20 30 40 50 60 70 80 1000 1 / [N O2] (M-1) Time (sec) B A Rate constant is the value of the slope (no sign) = 0.06904 Questions 3 and 4 refer to the following information: The rate constant of the decomposition of nitrogen dioxide was measured at various temperatures. The plot of ln(rate constant) versus 1/temperature is shown below and the line of best fit has been drawn through all data points. 3. The slope of the line can be calculated as (c) -1.378 x 104 K Slope = = = -13780 K 4. Given that R = 8.314 J K–1 mol–1 , the activation energy for this reaction is calculated as5.5 6 6.5 7 7.5 8 8.5 9 9.5 0.0015 0.00155 0.0016 0.00165 0.0017 ln(rate constant) 1/temperature (1/K) x = 0.0000900 K-1 y = -1.240 Questions 5 and 6 refer to the following information: The reaction of Br2(g) and NO(g) forms BrNO (g) according to the equation: Br2(g) + 2 NO(g) 2 BrNO(g) The following mechanism is proposed for this reaction: Step 1 Slow Br2 (g) + NO (g) Br2NO (g) Step 2 Fast Br2NO (g) + NO (g) 2 BrNO (g) 5. For this mechanism to be consistent with kinetic data, the experimental rate law would have been? (d) Rate = k [Br2] [NO] The rate law is defined by the “slow” step in the reaction. The rate law will be the product of the rate constant and the concentrations of the reactants in the slow step 6. The intermediate(s) in this proposed reaction mechanism is/are (c) Br2NO Species that are formed in the initial steps of the mechanism that are not products 7. What is the IUPAC name of the following molecule H C3 C H C H2 C O C H2 C O H C H Cl 2 CH3 CH3 (a) ethyl 4-chloro-3-methylhexanoate length of carbon chain 6 carbons = hex, no double or triple bonds = hexan, ftil(ildibl)tht5 6 4 3 2 1 8. Which of the carbon atom(s) labeled in the molecule below is/are chiral? C H C H CH3 C H H C2 C H2 C C H2 CH3 CH3 H C3 OH 1 2 3 (c) 2 only Chiral carbon is a carbon with four bonds, each bond connected to a different group, alkene are not chiral carbons (not 3) carbon 1 has two ethyl (CH2CH3) attached.