15. How many nearest neighbors surround each particle in a face-centered cubic lattice?
(a) 4 (b) 6 (c) 8 (d) 12
The picture above shows the face centered cubic lattice. On each side and vertex are the atoms.
In order to find how many nearest neighbors surround each particle, we need to combine the 8 lattices and make a bigger cube.
If we take the atom in the center of the larger cube, the nearest atoms will be the ones in the middle of the face of the lattice. If you draw a picture, you can easily get this. The answer will be (d) 12.
16. Hydrogen is collected over water at 22 ℃ and a barometer reading of 740 mmHg. If 300 ml of hydrogen is collected, which expression will give the volume of dry hydrogen at the same temperature and pressure?
Vapor Pressure at 22 ℃ : 20 mmHg
(a) 300ml × 740mmHg -20mmHg/740mmHg
(b) 300ml × 740mmHg + 20mmHg/740mmHg
(c) 300ml × 740mmHg/740mmHg-20mmHg
(d) 300ml × 740mmHg/740mmHg + 20mmHg
The point here is that we have to find the volume of hydrogen when the temperature is 22 and the pressure is 740 mmHg. Keep in mind that the pressure of hydrogen that is collected is 740 mmHg -20 mmHg (we have to subtract the vapor pressure of water). Therefore, what we need is the equation P1V1=P2V2. On the left side we plug in 720 mmHg × 300ml. On the right side we have 740 mmHg × V2 (the volume we are finding). Therefore, the answer is (a).
17. What is the normal melting point of the substance represented by the phase diagram?
(a) A (b) B (c) C (d) D
A is the triple point where the gas, solid, and liquid state coexist. Since we have to find the melting point, it should be in 1 atm. Therefore, we can eliminate C. Between B and D, B is at the border between the solid state and liquid state. So the answer is (b).
18. A bomb calorimeter has a heat capacity of 783 J·℃^-1 and contains 254 g of water, which has a specific heat of 4.184 J·g^-1·℃^-1 . How much heat is evolved or absorbed by a reaction when the temperature goes from 23.73℃ to 26.01℃?
(a) 1.78 kJ absorbed (b) 2.42 kJ absorbed (c) 1.78 kJ evolved (d) 4.21 kJ evolved
The temperature difference is 2.28℃. Therefore, the heat related to the calorimeter is 2.28 × 783 J.
The heat related to the water will be 4.184 × 2.28 × 254 J. The sum of the both is 4.21kJ. Therefore the answer is (d). However, in this problem I’m not sure why it should be heat evolved, not absorbed.
Edit: The heat is evolved because the heat from the reaction raised the temperature of the bomb calorimeter.
19. Consider this equation and the associated value for ΔH^°.
2H_2(g) + 2Cl_2(g) → 4HCl(g)
ΔH^° = -92.3kJ
Which statement about this information is incorrect?
(a) If the equation is reversed, the value ΔH^° equals 92.3kJ.
(b) The four HCl bonds are stronger than the four bonds in H_2 and Cl_2.
(c) The ΔH^° value will be -92.3kJ if the HCl is produced as a liquid.
(d) 23.1 kJ of heat will be evolved when 1 mol of HCl(g) is produced.
Let’s check all the choices step by step. The first choice is correct. The reverse reaction will be an endothermic reaction. The second choice is also correct. The fact that this reaction is an exothermic reaction means that the products are more stable than the reactants. We need energy to break the H_2 bonds and Cl_2 bonds but the energy that is released by forming the H-Cl bond is much greater that the reaction becomes exothermic. In other words, H-Cl bonds are much stronger than the H-H and Cl-Cl bonds. We can also see this from the difference in electronegativity. Now the third choice is incorrect because the reaction we have shows that the product is HCl gas not liquid. If it were liquid the value of ΔH^° will be different from -92.3kJ. The last choice is correct because we have -92.3kJ for four mols of HCl.
20. Determine the heat of reaction for this process.
FeO(s) + Fe_2O_3(s) → Fe_3O_4(s)
2Fe(s) + O_2(g) → 2FeO(s) ΔH° = -544.0kJ
4Fe(s) + 3O_2(g) → 2Fe_2O_3(s) ΔH° = -1648.4kJ
Fe_3O_4(s) → 3Fe(s) + 2O_2(g) ΔH°= + 1118.4kJ
(a) -1074.0kJ (b) -22.2kJ (c) +249.8kJ (d) +2214.6kJ
In order to solve this we need to use Hess’s Law.
Hess’s Law states that the enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps (Wikipedia).
In other words, by manipulating the given equations above we can determine the heat of reaction.
If we reverse the first and second reaction and divide them by 2 and then add them, you get the left side of the reaction that we want. Then, if you subtract the third reaction from the sum of the first and second reaction we have, you get the reaction that we are looking for.
Therefore, the answer is (b).
21. For which process will ΔH° and ΔG° be expected to be most similar?
(a) 2Al(s) + Fe_2O_3(s) → 2Fe(s) + Al_2O_3(s)
(b) 2Na(s) + 2H_2O(l) → 2NaOH(aq) + H_2(g)
(c) 2NO_2(g) → N_2O_4(g)
(d) 2H_2(g) + O_2(g) → 2H_2O(g)
For this problem, we should know the relationship between ΔH° and ΔG° which is G = H -TS in short. In order for the ΔH° and ΔG° to be similar, TS should be 0 at best (S is the entropy). In other words, we have to minimize TS as far as we can. If we look at the given choices, we can’t know about the temperature. So temperature doesn’t count. What matters is the entropy. From the reactions, we can’t know the exact amount of change in entropy but we can at least know if there is a change in the entropy. The answer will be (a) because there is relatively less change in entropy than other choices because there is no phase change and the sum of the mols of the reactants equals that of the products.
22. Use bond energies to estimate ΔH for this reaction.
H_2(g) + O_2(g) → H_2O_2(g)
Bond Bond Energy
H-H 436 kJ · mol^-1
O-O 142 kJ · mol^-1
O=O 499 kJ · mol^-1
H-O 460 kJ · mol^-1
(a) -127kJ (b) -209kJ (c) -484kJ (d) -841kJ
It takes energy to break the H-H bond and the O=O bond. Then as O-H bond and O-O bond are formed, energy is released. Therefore, we must subtract the total energy released as the new bonds form from the sum of the energy required to break the bonds of H_2 and O_2. The answer will be (a).
23. For a particular reaction, ΔH°= -38.3kJ and ΔS° = -119 J · K^-1. This reaction is
(a) spontaneous at all temperatures.
(b) nonspontaneous at all temperatures.
(c) spontaneous at temperatures below 66 ℃.
(d) spontaneous at temperatures above 66 ℃.
Here we use ΔG= ΔH – TΔS. Assuming that the pressure and temperature are stable, the forward reaction will occur spontaneously if ΔG is negative. In this problem we are given ΔH and ΔS. So what matters is the temperature. The value of ΔG will be negative in cases when the temperature is below 66 ℃. Therefore the answer is (c).
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