KEY: FREE RESPONSE. Question Ammonium carbamate, NH 4CO 2NH 2, reversibly decomposes into CO 2 and NH 3 as shown by the balanced equation below.

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1 KEY: FREE RESPONSE Question 1 1. Ammonium carbamate, NH 4CO 2NH 2, reversibly decomposes into CO 2 and NH 3 as shown by the balanced equation below. NH 4CO 2NH 2(s) 2 NH 3(g) + CO 2(g) K p =? A student places 0.50 g of solid NH 4CO 2NH 2 in a previously evacuated, rigid L container that is held at a constant temperature of 298 K. The student monitors the total pressure in the container as a function of time and obtains the graph shown below. (a) Based on the information in the graph and the reaction above, determine (i) the final partial pressure of NH 3(g) in the L container Based on the information in the graph, the total pressure at equilibrium is equal to 65 torr. Based on the 2-to-1 mole ratio between NH 3(g) and CO 2(g) in the balanced chemical equation, P NH3 = 2 (P CO2 ) answer with supporting work. (note units must be given here) P total = P NH3 + P CO2 = 2x + x = 3x = 65 torr P CO2 = x = (65 3) torr = torr P NH3 = 2x = (2)(21.7) torr = torr = 43 torr (sf) If P NH3 is expressed in units of atmospheres, P NH3 = atm

2 Question 1 (continued) (ii) the value of K p for the decomposition of NH 4CO 2NH 2 at 298 K. P NH3 = P CO2 = 1 point earned for correct Ksp expression 1 point earned for correct value based upon answer to a i) K p = ( atm) 2 ( atm) K p = (iii) the mass of solid NH 4CO 2NH 2 (molar mass g/mol) that decomposed 1 point is earned for correctly using the ideal gas law to calculate the number of moles of CO 2 (or moles of NH 3). = g NH 4CO 2NH 2 1 point is earned for the conversion of moles of CO 2 (or moles of NH 3) into grams of NH 4CO 2NH 2. The student is asked to perform the experiment a second time with one of the following changes to the experimental conditions. Perform the experiment at a higher temperature of 325 K. Perform the experiment with 1.00 g of NH 4CO 2NH 2. Perform the experiment in a 1.00 L container. Perform the experiment in a L container. (b) Which of the changes to the experimental conditions listed above will result in a greater mass of NH 4CO 2NH 2 decomposing than was calculated in part (a) (iii)? Justify your answer. The total pressure in the container is independent of the size of the container as long as solid remains. However, in a larger container, more moles of gas must form to maintain the 65 torr pressure. Hence, in the 1.00 L container, a greater amount of solid must decompose to reach equilibrium. 1 point for correct selection. 1 point for correct explanation based upon equilibrium principles. (note discussion of Delta H in webinar)

3 Question 1 (continued) Although it has not been directly observed, NH 2COOH has been proposed as an important intermediate in the decomposition of ammonium carbamate. (c) In the box below, complete the Lewis electron-dot diagram for NH 2COOH by drawing in the electron pairs. 1 point is earned for a correctly drawn diagram. (d) Based upon the Lewis diagram drawn above, what is the approximate O C O bond angle in NH 2COOH? 120 degrees 1 point is earned for an acceptable angle that is consistent with the Lewis diagram. NH 2COOH is predicted to be a highly soluble Brønsted-Lowry acid when placed in water. (e) Using principles of intermolecular forces, explain why NH 2COOH would be very soluble in water. NH 2COOH is soluble in water because it can form multiple hydrogen bonding interactions with water molecules. Therefore the energy required to separate one water molecule from another is comparable to the energy of interaction of NH 2COOH with water. Hence the NH 2COOH dissolves. 1 point is earned for a valid explanation. (f) Circle the atom in the diagram above that will be released when NH 2COOH acts as an acid in aqueous solution. See the diagram above. The hydrogen atom connected to the oxygen atom in the diagram should be circled. 1 point is earned for circling the correct hydrogen atom in the diagram.

4 Question 2 2 O 3(g) 3 O 2(g) ΔH o = 286 kj/mol rxn 2. Ozone, O 3, reacts to form O 2 as shown by the equation above. (a) If 4.0 moles of O 3(g) completely react to form O 2(g), calculate the magnitude of heat energy that is released or absorbed. 4.0 mol O + -./ / :; -./0 123 = -570 kj answer with supporting work. The rate law for the reaction is known to be the following. rate = k[o 3] (b) The half-life of the reaction is 31 minutes at 25 o C. Determine the rate constant for the reaction. Please include units in your answer. Based on first-order kinetics, = min 1 value of k with the correct units. (c) A student places a M sample of O 3(g) in a previously evacuated reaction vessel. The student then monitors the production of O 2(g) as the reaction above proceeds at a constant temperature of 25 o C. Calculate the concentration of O 2(g) produced 45 minutes after the reaction has started. Based on first-order kinetics, 1 point for correct calculation of O 3 after 45 minutes M O 3 remains 1 point for correct calculation of O 2 produced after 45 minutes [O 3] reacted = = * 2mol O 2 / 3 mol O 3 = M O 2 produced

5 Question 3 3. Answer the following questions about the elements nitrogen, fluorine, and the ions they form. Atom First Ionization Energy (kj mol 1 ) F 1681 N 1402 (a) Using principles of atomic structure, explain why the first ionization energy of atomic fluorine is greater than that of atomic nitrogen. The valence electrons of both nitrogen and fluorine are located in the same subshell (2p). These electrons experience forces of repulsion from core as well as other valence electrons. However, the core repulsion is identical while the repulsion of one valence electron on another has a minimal effect. However, fluorine has more protons in its nucleus than nitrogen does. Therefore, the electrons in the 2p sublevel in fluorine experience a greater attraction than those in N. The greater attraction with effectively the same overall forces of repulsion among the electrons make it harder to pull an electron away from F, making its first ionization energy higher. 1 point is earned for a valid explanation. The common anions of atomic nitrogen and atomic fluorine (the nitride and fluoride anions) that are found in ionic compounds are isoelectronic to one another. (b) Pick one of the elements above. Write the formula of the most common ion it would form. Write the complete electron configuration of that ion. N 3 or F 1s 2 2s 2 2p 6 formula and charge of the anion. electron configuration. (c) Which of the two ions has the greater size? Justify your answer using principles of atomic structure. N 3 is larger in size than F. Both anions have the same electron configuration. Therefore all forces of repulsion among electrons in both species are the same. However, F has more protons than N 3. Therefore the electrons in the F ion experience a stronger attraction toward the nucleus than those of the N 3 ion. This causes the radius of F to be smaller than that of N 3. choice with a valid justification.