USNCO Problems 16

24. What is ΔG° for this reaction?

1/2N_2(g) + 3/2H_2(g) ⇔ NH_3(g)        K_p= 4.42 × 10^4 at 25℃.

(a) -26.5 kJ · mol^-1 (b) -11.5 kJ · mol^-1 (c) -2.2 kJ · mol^-1 (d) -0.97kJ · mol^-1

Here we have to use this equation ΔG°= -RT lnK which shows the relationship between ΔG° and the equilibrium constant. The answer will be (a).

25. A reaction follows this concentration-time diagram. The instantaneous rate for this reaction at 20 seconds will be closed to which value?

(a) 4 × 10^-3 M · sec^-1 (b) 8 × 10^-3 M · sec^-1 (c) 2 ×10^-2 M · sec^-1 (d) 1 × 10^-1 M · sec^-1

Here the key word is the “instantaneous rate.” The rate will be the Δmolarity/Δtime. However, as we are to find the instantaneous rate, we need to find the slope of the line that is tangent to the parabola at 20 s. If  we were given the equation of parabola, we could easily take the differential of the parabola. But here it is easy to find the slope which will be 0.20/40s = 5 × 10^-3 M. Therefore, the answer is (a).

26. If the half-life of a reaction increases as the initial concentration of substance increases, the order of the reaction is

(a) zero. (b) first. (c) second. (d) third.

First of all the answer is (a). The equation for half-lives for each order will be

t_½ =  [A_o] / 2k for the zero order,

t_½ = 0.693 / k for the first order,

t_½ = 1 / k [A_o] for the third order.

As we can see, in the first order reaction, the half-life has nothing to do with the initial concentration. Therefore, as we can see from the above equations, the answer will be the zero order.

27. The radioisotope N-13, which has a half-life of 10 minutes, is used to image organs in the body. If an injected sample has an activity of 40 microcuries, what is its activity after 25 minutes in the body?

(a) 0.75 µCi (b) 3.5 µCi (c) 7.1 µCi (d) 12 µCi

Half life means the time it takes for a substance to become half of its initial concentration or amount. For example, if the half time is 2 minutes, after 2 minutes the substance will be 4 g from 8 g. In this case, the half-life is 10 minutes. Therefore, after 20 minutes N-13′s activity will 1/4 of 40. After another 5 minutes, the activity will decrease again from 10. However, it would be greater than 5 because only 5 minutes, not 10 minutes, have passed. Therefore, we can deduce that the activity will be somewhere between 5 and 10. So the answer is (c).

28. Propanone reacts with iodine in acid solution as shown in this equation.

H^+

CH_C(O)CH_3 + I_2  →  CH_3C(O)CH_2I +HI

These data were obtained when the reaction was studied.

What is the rate equation for the reaction?

(a) rate = k[CH_3C(O)CH_3][I_2]

(b) rate = k[CH_3C(O)CH_3]^2

(c) rate = k[CH_3C(O)CH_3][I_2][H^+]

(d) rate = k[CH_3C(O)CH_3][H^+]

In order to determine the rate equation, you have to use the given data. When everything else stays the same, if we double the concentration of CH_3C(O)CH_3, the rate also doubles. Therefore, it will be first order for CH_3C(O)CH_3. If the concentration of I_2 doubles (when every other factor stays the same), there is no change in the rate. Therefore, it will be zero order for I_2. Lastly, when the concentration of H^+ is doubled the rate also doubles. Therefore, it will be first order. Therefore the answer is (d).

USNCO Problems 15

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?

Compound H_2O

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)

Given information:

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).

USNCO Problems 14

8. Methyl-t-butyl ether, C_5H_12O, is added to gasoline to promote cleaner burning. How many moles of oxygen gas, O_2, are required to burn 1.0 mol of this compound completely to form carbon dioxide and water?

 

(a) 4.5 mol (b) 6.0 mol (c) 7.5 mol (d) 8.0 mol

 

First thing you have to do when you see these kinds of problems is that you have to write the full balanced equation which will be

 

2C_5H12O + 15O_2  → 10CO_2 + 12H_2O

 

We see that 2 mols of C_5H_12O reacts with 15mols of O_2. Therefore the answer will be (c).

9. A 2.00g sample of benzoic acid, C_6H_5COOH (molar mass 122.1 g/mol), is titrated with a 0.120 M Ba(OH)_2 solution. What volume of the Ba(OH)_2 solution is required to reach the equivalence point?

(a) 6.82 ml (b) 13.6 ml (c) 17.6 ml (d) 35.2 ml

 

When you have titration problems, you should match the mols of acid and base that react. First you have to check if the acid is polyprotic or the base is polyacidic. In this case the base is polyacidic and can give 2 OH^- ions. Therefore, 2 mols of benzoic acid will react with 1 mol of Ba(OH)_2.

 

We can determine how many mols of benzoic acid we have since we know the molar mass and the amount existing–we have 0.001638 mols of benzoic acid.

 

Then we will need 0.000819 mols of Ba(OH)_2. The rest is easy. 0.000819 mols = 0.120M × the volume we need to find. Therefore the answer will be (a).

10. Chlorine can be prepared by reacting HCl with MnO_2. The reaction is represented by this equation.

MnO_2(s) + 4HCl(aq)  → Cl_2(g) + MnCl_2(aq) + 2H_2O (l)

Assuming that the reaction goes to completion what mass of concentrated HCl solution (36.0% HCl by mass) is needed to produce 2.50g of Cl_2?

(a) 5.15g (b) 14.3g (c) 19.4g (d) 26.4g

 

First of all you should find out that we need 4 mols of HCl to produce 1 mol of Cl_2. Then you have to determine how many mols will 2.50g of Cl_2 will be. 2.50g of Cl_2 will be 0.0353mols. Therefore we need 0.141 mols of HCl and that will be 5.14g of HCl. However, we have to find the mass of the HCl solution not just HCl.

 

Since we know that the concentration is 36%, we can easily find the mass of the solution which is 14.3g. The answer is (b).

11. What is the Na^+ ion concentration in the solution formed by mixing 20.ml of 0.10 M Na_2SO_4 solution with 50.ml of 0.30M Na_3PO_4 solution?

(a) 0.15 M (b) 0.24 M (c) 0.48 M (d) 0.70 M

 

You should first determine the sum of the Na^+ mols that you get from each solutions. You’ll get 0.02 L × 0.1 M × 2 + 0.05 L × 0.30 M × 3 mols of Na^+. Since it’s asking for the molar concentration you divide the number of mols by the total volume which is 0.07 L. The answer is (d).

12.

A solution prepared by dissolving a 2.50 g sample of an unknown compound dissolved in 34.0g of benzene, C_6H_6, boils 1.38 ℃ higher than pure benzene. Which expression gives the molar mass of the unknown compound?

 

K_b : 2.53  ℃  × m^-1

 

(a) 2.53 × 2.50/1.38

(b) 1.38 × 34.0/2.53 × 2.50

(c) 2.50 ×10^3 × 2.53/34.0 × 1/1.38

(d) 2.50 × 10^3 × 1.38/34.0 × 2.53

 

Now the boiling point elevation and freezing point depression depends on the molality of the solution.

Therefore 1.38℃  = 2.50g/the molar mass of the unknown × 1/0.034Kg of benzene × 2.53 ℃ × m^-1

 

 

The answer will be (c).

13. What is the total pressure in a 2.00L container that holds 1.00g He, 14.0g CO, and 10.0g of NO at 27.0 ℃ ?

 

(a) 21.6 atm (b) 13.2 atm (c) 1.24 atm (d) 0.310 atm

 

Okay the total pressure equals the sum of the partial pressure of the gases. Therefore, you must first find the partial pressure of each gas. We have the volume, mol, temperature. So we can use the ideal gas equation to find the partial pressure of each gas. It’s simple The answer will be (b).

14. What type of solid is generally characterized by having low melting point and low electrical conductivity?

(a) ionic (b) metallic (c) molecular (d) network covalent

 

We can first eliminate (a) because ionic solids have high melting points. We can also eliminate (b) because metals have high electrical conductivity. Therefore, the answer is (c) which have relatively low melting points and low electrical conductivity. For (d), though it has low electrical conductivity it has high melting point.

USNCO Problems 13

USNCO 2000 Multiple Choice Questions

1. Which of these ions is expected to be colored in aqueous solution?

I Fe^3+  II Ni^2+  III Al^3+

 

(a) I only (b) III only (c) I and II only (d) I,II, and III

 

The answer will be (c) because I and II are transition metal ions. Transition metal ions are usually colored in aqueous solutions because their d-orbitals are filled in. The reason why transition metals have multiple oxidation number is also because of d-orbitals.

2. Which substance is stored in contact with water to prevent it from reacting with air?

(a) bromine

(b) lithium

(c) mercury

(d) phosphorus

 

Well the answer is phosphorus. Since phosphorus is insoluble in water, it is stored in water to prevent other reactions. Bromine should be stored in glass bottles. Lithium is stored in organic solvents.

3. A solution of concentrated aqueous ammonia is added dropwise to 1ml of a dilute aqueous solution of copper(II) nitrate until a total of 1ml of the ammonia solution has been added. What observations can be made during this process?

 

(a) The colorless copper(II) nitrate solution turns blue and yields a dark blue precipitate.

(b) The colorless copper(II) nitrate solution yields a white precipitate which turns dark blue upon standing.

(c) The light blue copper(II) nitrate solution yields a precipitate which redissolves to form a dark blue solution.

(d) The light blue copper(II) nitrate solution turns dark blue and yields a dark blue precipitate.

 

First of all you can see that (a) and (b) cannot be the answers because the aqueous solution of copper(II) nitrate is blue. Next, let’s see the total reaction which is

 

[Cu(H₂O)₄]²⁺ + 4NH₃ = [Cu(NH₃)₄]²⁺ + 4H₂O

It’s because copper(II) nitrate solution exists as the copper hexaaqua complex ion and free nitrate ions. The answer is (c). The dark blue color is due to [Cu(NH₃)₄]²⁺ . Since I never tried out this reaction, I don’t know about the precipitate but I assume the precipitate which redissolves will be Cu(OH)_2

4. What gas is produced when dilute HNO_3 is added to silver metal?

(a) NO (b) H_2 (c) NH_3 (d) N_2

 

You might choose (b) H_2 as it’s a reaction between an acid and metal. However, because silver’s reactivity is lower than that of hydrogen, hydrogen will not be reduced. Therefore, we should try to write the balanced equation.

 

However, keep in mind that we are talking about dilute nitric acid here.

 

The equation for the concentrated nitric acid and silver will be

2HNO_3 + Ag ↔ NO_2 + AgNO_3 + H_2O

 

But the equation for the dilute nitric acid will be

 

4HNO_3 + 3Ag ↔ NO + 3AgNO_3 +2H_2O

 

Therefore the answer will be (a).

 

And here is a link to a site about nitric acid. http://www.chemicalland21.com/industrialchem/inorganic/nitric%20acid.htm

5. A substance is analyzed by paper chromatography, giving the chromatogram shown.

 

(a) 0.80 (b) 0.75 (c) 0.67 (d) 0.60

 

The retardation factor (R_f) is defined as the ratio of the distance traveled by the center of a spot to the distance traveled by the solvent front (Wikipedia). Therefore, the answer will be (b).

6. The molarity of a Cu^2+ solution is to be determined from its absorbance, measured under the same conditions as those used to prepare this calibration curve. What will be the percent uncertainty in the concentration of a 0.050M solution if the uncertainty in the absorbance reading is ±0.01 absorbance units?

 

 

 

 

 

(a) 5 % (b) 10 % (c) 15 % (d) 20 %

 

As you can see from the graph the absorbance will be 0.10 for 0.050M of Cu^2+. The percent uncertainty is just the uncertainty over the actual value multiplied by 100. Therefore the answer will be 0.01/0.10 × 100 = 10%, (b).

 

7. A 1.50g sample of an ore containing silver was dissolved, and all of the Ag^+ was converted to 0.124g of Ag_2S. What was the percentage of silver in the ore?

(a) 6.41% (b) 7.20% (c) 8.27% (d) 10.8%

 

This is an easy simple question. First of all you have to determine how many grams of silver are in 0.124g of Ag_2S by using the mass percentage of silver in Ag_2S. After you find out the mass of silver in 0.124 which will be about 0.108g, you just have to calculate what percentage will 0.108g will be from 1.5g. The answer is (b).

 

MO Diagram for F2

MO Diagram for HF

MO Diagram for Hydrogen