PORTAGE LEARNING CHEM 104

PORTAGE LEARNING CHEM 104 Module 1: Problem Set Due No due date Points 0 Questions 18 Time Limit None Take the Quiz Again Question 1 Not yet graded / 0 pts In the reaction of 0.200 M gaseous N2O5 to yield NO2 gas and O2 gas as shown below: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) the following data table is obtained: Time (sec) [N2O5] [O2] Attempt History Attempt Time Score LATEST 2O22 Attempt 1 6 minutes * Some questions not yet graded Score for this quiz: 0 out of 0 * Submitted Jun 17 at 2:59pm This attempt took 6 minutes. 0 0.200 M 0 300 0.180 M 0.010 M 600 0.019 M   900 0.146 M 0.027 M 1200 0.132 M 0.034 M 1800 0.110 M 0.045 M 2400 0.096 M 0.052 M 3000 0.092 M 0.054 M (a) Use the [O2] data from the table to calculate the average rate over the measured time interval from 0 to 3000 secs. (b) Use the [O2] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 300 sec). (c) Use the [O2] data from the table to calculate the instantaneous rate late in the reaction (2400 secs to 3000 secs). (d) Explain the relative values of the average rate, early instantaneous rate and late instantaneous rate. Your Answer: . Question 2 Not yet graded / 0 pts In the reaction of gaseous CH3CHO to yield CH4 gas and CO gas as shown below: CH3CHO (g) → CH4 (g) + CO (g) the following data table is obtained: 0 0.0500 M 1200 0.0300 M 2000 0.0240 M 6000 0.0120 M 10,000 0.0080 M 15,000 0.0056 M 20,000 0.0043 M Time (sec) [CH3CHO] (a) Use the [CH3CHO] data from the table to calculate the average rate over the measured time interval from 0 to 20,000 secs. (b) Use the [CH3CHO] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 1200 sec). (c) Use the [CH3CHO] data from the table to calculate the instantaneous rate late in the reaction (15,000 secs to 20,000 secs). (d) Explain the relative values of the average rate, early instantaneous rate and late instantaneous rate. Your Answer: . Question 3 Not yet graded / 0 pts The following rate data was obtained for the reaction which takes place in a solution of OH-: ClO- + I- → IO- + Cl- Experiment # [I-] [ClO-] [OH-] rate 1 0.0030 0.0010 1.00 1.8 x 10-4 2 0.0030 0.0020 1.00 3.6 x 10-4 3 0.0060 0.0020 1.00 7.2 x 10-4 4 0.0030 0.0010 0.50 9.0 x 10-5 Determine the reaction order with respect to (1) ClO-, (2) I- and (3) OH- Your Answer: .   Question 4 Not yet graded / 0 pts Following the reaction order that was determined in the previous questions, answer these two parts: (4) write the rate law and then (5) determine the value of the rate constant, k. Your Answer: . Question 6 Not yet graded / 0 pts Determine the decay constant (k) and the half-life of a radioactive nucleus if 75% of the material has decayed in 400 days. Your Answer: . Question 7 Not yet graded / 0 pts A sample of wood from an ancient tomb was found to contain 15.7 % 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 yrs. What is the age of the wood? Your Answer: . Question 8 Not yet graded / 0 pts Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4. CO (g) + 3 H2 (g) CH4 (g) + H2O (g) Your Answer: . Not yet graded / 0 pts Question 9 Calculate the Kc for the following reaction if an initial reaction mixture of 0.700 mole of NH3 and 0.910 mole of O2 in a 1.00 liter container forms an equilibrium mixture containing 0.230 mole of N2 and corresponding amounts of NH3, O2, and H2O. Question 10 Not yet graded / 0 pts Question 11 Not yet graded / 0 pts The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants. Be sure to explain your answer. N2 (g) + O2 (g) 2 NO (g) Kc = 4.6 x 10-31 Your Answer: . Question 12 Not yet graded / 0 pts Explain the following: (a) Explain why reaction rate is affected by a reactant concentration change. (b) Explain why reaction rate is affected by a temperature change. (c) Explain why powdered zinc reacts faster than a chunk of zinc. (d) Explain how a catalyst works. (e) Explain why reactions in solution are usually faster than heterogeneous reactions. (f) Explain why stirring speeds up the rate of heterogeneous reactions. Your Answer: . (a) A change in reactant concentration changes the number of particles colliding which changes the probability of the particles undergoing the reaction. (b) As the temperature is changes, the kinetic energy of the reacting particles is changed and this changes the probability of particles colliding with sufficient energy to undergo the reaction. (c) Solid particles only react at their surface and since powdered zinc particles have a greater surface energy than a chunk of zinc the powdered zinc will react faster. (d) A catalyst speeds up the rate of a chemical reaction by causing formation of an alternate transition state with lower Energy of Activation, thereby allowing more reactant molecules to form the Transition State at lower energy. (e) The reactants particles can collide with one another much more readily in solution whereas particles in heterogeneous reactions can only collide by way of their surfaces. An increase in collisions causes an increase in reaction rate. (f) Stirring heterogeneous reactions increases the collisions between particles which speeds up the rate of reaction. Question 13 Not yet graded / 0 pts The equilibrium reaction below has the following equilibrium mixture concentrations: SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] with Kc = 4.00 If the concentration of O2 at equilibrium is decreased to [0.100], how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2 SO3 (g) 2 SO2 (g) + O2 (g) Your Answer: Question 14 Not yet graded / 0 pts The equilibrium reaction below has the following equilibrium mixture concentrations: SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] with Kc = 4.00 If the concentration of SO3 at equilibrium is increased to [0.300], how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2 SO3 (g) 2 SO2 (g) + O2 (g) Your Answer: . Question 15 Not yet graded / 0 pts The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is decreased from 4.00 liters to 2.00 liters, how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2 SO3 (g) 2 SO2 (g) + O2 (g) Your Answer: . Question 16 Not yet graded / 0 pts The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is increased from 3.00 liters to 9.00 liters, how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2 SO3 (g) 2 SO2 (g) + O2 (g) Your Answer: . Question 17 Not yet graded / 0 pts The equilibrium reaction below has the Kc = 4.00 at 25oC. If the temperature of the system at equilibrium is increased to 100oC, how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change. 2 SO3 (g) 2 SO2 (g) + O2 (g) ∆H0 = + 94.1 kJ Your Answer: . If the temperature is increased on this reaction at equilibrium which has a +∆H0, the reaction will briefly shift in the direction that uses up some of the added heat [+∆H0 indicates an endothermic (heat absorbing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of SO2 and O2 (in the numerator of Kc) and decrease the concentration of SO3 (in the denominator of Kc) which will increase the value of Kc. Question 18 Not yet graded / 0 pts The equilibrium reaction below has the Kc = 0.250 at 25oC. If the temperature of the system at equilibrium is decreased to 0oC, how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change. 2 SO2 (g) + O2 (g) 2 SO3 (g) ∆H0 = - 94.1 kJ Your Answer: . If the temperature is decreased on this reaction at equilibrium which has a -∆H0, the reaction will briefly shift in the direction that produces some heat [-∆H0 indicates an exothermic (heat producing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of SO3 (in the numerator of Kc) and decrease the concentration of SO2 and O2 (in the denominator of Kc) which will increase the value of Kc. Quiz Score: 0 out of 0