Spring 2004 BCHS 3304 Practice Exam I-1). An enzyme (protein catalyst) catalyzes the reaction: A B once per microsecond. How many times does the enzyme catalyze the reaction in one minute?a). 1 X 108 min-1b). 6 X 108 min-1c). 1 X 107 min-1d). 6 X 107 min-1e). 6 X 109 min-1(d), 1 X 106 X 60 = 6 X 107 min-1.2). A DNA helix is 2 nanometers in diameter. How many DNA helices can you fit side-by-side in 1 decimeter?a). 5 X 108 helices b). 2 X 107 helicesc). 5 X 10-7 helices d). 2 X 109 helicese). 5 X 107 helices(e), 0.1 meters / 2 X 10-9 meters per helix = 5 X 107 helices.3). When 1 mole of crystalline sucrose is placed in 1 L pure water, the reaction vessel gets cool to the touch, and it takes many minutes for all of the sucrose to dissolve at room temperature. Given what you know of thermodynamics, what are the signs of the values of the change in enthalpy and entropy of the system?a). H +; S - b). H +; S +c). H -; S - d). H -; S +(b), H +; S +.4). A chemical reaction occurs in a beaker of water. The internal energy of the system did not change throughout the course of the reaction, but the system lost 4 kJ of heat to the surroundings. What does this imply about the work done by/on the system?a). The surroundings did 4 kJ of work on the system.b). The system did 4 kJ of work on the surroundings.c). This is not possible as it violates the 1st law of Thermodynamics.d). This is not possible as it violates the 2nd law of Thermodynamics.e). Cannot calculate due to insufficient information.(a), the surroundings did 4 kJ of work on the system.5). A process has the following thermodynamic parameters: H’ is “-“ and S’ is “-“. At what temperatures will this process be spontaneous?a). Spontaneous at temperatures below T = H’/S’.b). Spontaneous at temperatures above T = H’/S’.c). Spontaneous at all temperatures.d). Not spontaneous at any temperature.(a), spontaneous at temperatures below T = H’/S’.6). The hydrolysis of phosphoenolpyruvate to pyruvate and inorganic phosphate (Pi) is represented by the following reaction and has the indicated G’:H2O + phosphoenolpyruvate  pyruvate + Pi + H+ G’ = -61.9 kJ mol-1The hydrolysis of ATP is represented by the following reaction and has the indicated G’:H2O + ATP  ADP + Pi + H+G’ = -30.5 kJ mol-1Write the net chemical equation for the synthesis of ATP from ADP and inorganic phosphate using the hydrolysis of phosphoenolpyruvate to drive the reaction. Calculate the net G’ for the overall reaction and indicate whether the overall reaction would be spontaneous or non-spontaneous. H2O + phosphoenolpyruvate  pyruvate + Pi + H+G’ = -61.9 kJ mol-1 ADP + Pi + H +  ATP + H2O  G  ’ = +30.5 kJ mol -1 Phosphoenolpyruvate + ADP  pyruvate + ATP G’ = -31.4 kJ mol-1 The net, overall reaction is spontaneous.7). Given the following four types of chemical interactions: van der Waals bonds, ionic bonds, covalent bonds, and hydrogen bonds, list them in order from strongest to weakest in terms of enthalpy (H), and tell which are the most sensitive to the dielectric constant of water.a). Ionic bonds, covalent bonds, van der Waals bonds, hydrogen bonds; van der Waals and ionic bonds sensitive to water dielectric. b). Covalent bonds, ionic bonds, hydrogen bonds, van der Waals bonds; hydrogen bonds and ionic bonds sensitive to water dielectric.c). Covalent bonds, hydrogen bonds, ionic bonds, van der Waals bonds; hydrogen bonds and van der Waals bonds sensitive to water dielectric.d). van der Waals bonds, hydrogen bonds, ionic bonds, covalent bonds; van der Waals and covalent bonds sensitive to water dielectric.(b), covalent bonds, ionic bonds, hydrogen bonds, van der Waals bonds; hydrogen bonds and ionic bonds are the most sensitive to the dielectric constant of water.8). The equilibrium constant (K’eq) for the dissociation of a weak acid in water = 1.763 X 102. Given that the universal gas constant (R) = 8.3 X 10-3 kJ mol-1 K-1, calculate the G’ at room temperature (25C) for this acid dissociation.a). –12.79 kJ mol-1b). +12.79 kJ mol-1c). –1.07 kJ mol-1d). +1.07 kJ mol-1e). –5.55 kJ mol-1(a), –12.79 kJ mol-1.9). How many ml of 0.3 M HCl do you need to lower the pH of 1 liter of water from pH = 5 to pH = 4? You must show all work to receive full credit.pH = 4 is 1 X 10-4 moles H+.pH = 5 is 1 X 10-5 moles H+.We need to add 1 X 10-4 moles H+ – 1 X 10-5 moles H+ = 9 X 10-5 moles H+.9 X 10-5 moles / 0.3 moles liter-1 = 0.003 liters (3 ml) of 0.3 M HCl.10). What conjugate base to weak acid ratio is required to produce a pH = 4 acetate buffer (An acetate buffer is made from acetic acid {pK = 4.76} and acetate anion)? pH = pK + log([A-]/[HA); 4.0-4.76 = log([A-]/[HA); ([A-]/[HA) = 0.17, so17/100.11). What is the pH of a 0.1 M NH4+Cl- solution where the weak acid (pK = 9.25) dissociates 40%? You must show all work to receive full credit.NH4+Cl-  NH4+ + Cl-pH = pK + log([A-]/[HA]) 0.1 M 0.1 M 0.1 MNH4+  NH3 + H+pH = 9.25 + log(40/60)0.1 M 40% pH = 9.0712). A student prepared a solution which contains equal equivalents of succinic acid and sodium succinate and then discovered that the laboratory’s pH meter was broken. How could the student best estimate the pH of the solution?The pH of the solution would be equal to the pK of succinic acid, since [succinic acid] = [sodium succinate]; [A-] = [HA].pH = pK + log ([A-]/[HA]), pH = pK + log 1; log 1 = 0 so pH = pK.13). What is the pK of a weak acid when the acid to conjugate base ratio = 6.3, and gives a pH 4.63?a). 3.83 b). 4.47c). 4.74 c). 3.38e). 5.42pH = pK + log ([A-]/[HA]), 4.63 = pK + log (1/6.3) = (e), 5.4214). At what range does histidine (pKR = 6.04) serve as a good buffer and what is the ratio of histidine to H+at the lower pH limit of the buffering capability of histidine?a). pH range = 6.04-8.04; 10:1 b). pH range = 6.04-7.04; 1:10c). pH range = 5.04-7.04; 1:10 d). pH range = 5.04-6.04; 10:1e). pH range = 5.54-6.54; 1:10(c), pH range = 5.04-7.04; 1:10.15). What is the pH of a 0.1 M H3PO4 solution (assume that H3PO4 dissociates 100% through each deprotonation step).a). 0.70 b). 0.52c). 1.0 d). 0.39e). 0.31H3PO4  H+ + H2PO4-  H+ + HPO42-  H+ + PO43- 0.1 M