Gen Chem Class 4 Amplifire 1/2 and 2/2

A high pressure gaseous synthesis of hydrazine (H4N2) is governed by the following equilibrium: 2 NH₃ ↔ H₄N₂ + H₂, Keq = 10^-7 If a 1 L system is charged with 10 mol of NH3, how much hydrazine will be present at equilibrium? Click here to view the periodic table. ANSWER INCORRECT THE CORRECT ANSWER 0.003 mol YOU WERE SURE AND INCORRECT 0.01 mol 0.001 mol 0.03 mol

0.003 mol of H4N2 will be present at equilibrium. At equilibrium we can say that K = [H₄N₂][H₂]/[NH₃]² = 10-7. Allowing [H₄N₂] = [H₂] = x while [NH₃] = 10 - 2x, we get x2/(10 - 2x)2 = 10-7. As Keq is small we can safely disregard 2x in the denominator, giving x2 = 10-5, or x = 10-2.5. This value is between 10^-2 and 10^-3, meaning 0.003 is the only possible answer.

A scientist adds reactants and an enzyme to a beaker placed on the benchtop. As the reaction proceeds, which of the following events would have the smallest effect on the rate of the forward reaction? ANSWER INCORRECT The temperature of the beaker is changed. THE CORRECT ANSWER Products are removed from the beaker YOU WERE SURE AND INCORRECT The solution is stirred. More reactants are added to the beaker.

Removing products from the beaker would have the smallest effect on the rate of the forward reaction. Temperature, reactant concentrations, and activation energy are the three major factors that play a role in a reaction's kinetics. If the temperature of the beaker changes, the rate of the forward reaction will be affected. When more reactants are added or the mixture stirred, the probability of collisions among the molecules increases. Thus, the forward reaction rate will increase. While removing products will shift the equation to the right (favoring the production of more products) by Le Chatelier's Principle, this is an effect on the *equilibrium*, or thermodynamics, of the reaction and does not affect the rate of the forward reaction.

The formation constant (Kf) for the complex ion [Fe(en)3]2+ is 5.0 x 109, whereas the Kf for [Fe(ox)3]4- is 1.7 x 105 (en = 1,2 diaminoethane, ox = oxalate anion). Based on this information, which of the following statements is true? ANSWER INCORRECT YOU WERE SURE AND INCORRECT Complexes with bidentate ligands always have higher Kf values than the corresponding monodentate complexes. The higher the oxidation state on the transition metal, the larger the formation constant. THE CORRECT ANSWER 1,2-Diaminoethane is a stronger ligand with iron(II) than the oxalate ion. A reaction between oxalate ion and [Fe(en)3]2+ will form [Fe(ox)3]4- completely.

1,2-diaminoethane is a stronger ligand with iron(II) than the oxalate ion. Formation constants are equilibrium constants for the formation of complex ions. The larger the equilibrium constant of a reaction, the larger the ratio of products to reactants at equilibrium will be. The oxalate complex with iron(II) has a lower Kf than the diamine complex. Therefore, any competitive reaction will favor the diamine complex, so [Fe(en)3]2+ is favored to form, not [Fe(ox)3]4-. Since both complexes have iron(II) as the central metal, there isn't any information to support the relationship suggested between oxidation state and the value of the formation constant. When comparing competing ligands, the one with the larger Kf will yield the more stable complex, which is due to the strongest Lewis acid/Lewis base interactions. Therefore, the diamine must be the stronger ligand. Both of the ligands in the question are bidentate, so there is no information to use for comparison between ligand type and value of Kf.

The [Pb2+] of a solution is 1.25 × 10-3 M. Calculate the [SO42-] that must be exceeded before PbSO4 can precipitate. (The solubility product of PbSO4 at 25°C is 1.6 × 10-8.) Click here to view the periodic table. ANSWER INCORRECT YOU WERE SURE AND INCORRECT 1.3 × 10-4 M 1.6 × 10-8 M 6.4 × 10-6 M THE CORRECT ANSWER 1.3 × 10-5 M

1.3 × 10-5 M The concentrations of the dissolved ions must exceed the Ksp value for precipitate to form. Given the dissolution of PbSO4, PbSO₄(s) Pb2+(aq) + SO₄2-(aq) we have Ksp = [Pb2+][SO42-]. Therefore, the minimum concentration of SO42- needed to form precipitate is [SO42-] = Ksp/[Pb2+] = (1.6 × 10-8)/(1.25 × 10-3) = 1.3 × 10-5 M.

The equilibrium constant for the reaction H₂(g) + I₂(g) ↔2HI(g) is 617 at a given temperature. In a reaction vessel, the partial pressures of H2 and I2 are each 1 atm and the partial pressure of HI is 12 atm. What is the value of the reaction quotient, Q? ANSWER INCORRECT 7.0 × 10-3 THE CORRECT ANSWER 144 YOU WERE SURE AND INCORRECT 12 8.3 ×10-2

144. Although calculated in the same manner, Q is independent of Keq. Q can be solved using partial pressures Q=(P HI)²/ (P H₂) (P I₂) = (12)²/(1)(1) = 144

2 SO2(g) + O2(g) 2 SO3(g) A chemist adds a reagent to the following equilibrated reaction and immediately observes the forward and reverse reaction rates to increase by 2.14 and 2.29, respectively. Which of the following was most likely added to the reaction? Click here to view the periodic table. ANSWER V2O5(s) SO2(g) O2(g) SO3(g)

2 SO2(g) + O2(g) ↔ 2 SO3(g) Answer: V₂O₅ The situation here describes the addition of a catalyst to an equilibrated reaction. Catalysts increase the rates of both the forward and reverse reaction by decreasing the activation energy. Addition of a reactant (SO2(g) and O2(g)) would initially increase the rate of the forward reaction alone while adding a product (SO3(g)) would initially increase only the rate of the reverse reaction. Thus V2O5(s) must serve as a catalyst in this reaction and results in the observed rate changes.

Given the average rate for the following reaction is 2.25 × 10-2 M/min, how long will it take to generate 0.45 moles of NO2(g) in one liter of solution? 2 N2O5(g) → 4 NO2(g) + O2(g) Click here to view the periodic table. ANSWER INCORRECT YOU WERE SURE AND INCORRECT 20 minutes THE CORRECT ANSWER 5 minutes 10 minutes 15 minutes

Given the average rate for the following reaction is 2.25 × 10-2 M/min, it would take 5 minutes to generate 0.45 moles of NO2(g) in one liter of solution. The average rate expression for this reaction is as follows: rate=-1/2 [∆N₂O₅]/∆t = 1/4 ∆[NO₂]/∆t = ∆[O₂]/∆t = 2.25 x 10^-2 M/min Thus the change in nitrogen dioxide concentration over time is 0.09 M/min (four times that of the average rate) and 0.45 moles would form in 5 minutes (0.45 moles / 0.09 M/min).

The molar solubility of HgCl2 in water is 0.27 M. If 0.1 mol of NaCl is added to 1 L of a 0.27 M aqueous solution of HgCl2, what will happen? ANSWER INCORRECT THE CORRECT ANSWER NaCl will dissolve and HgCl2 will precipitate. NaCl will not dissolve and no precipitate will form. YOU WERE SURE AND INCORRECT NaCl will not dissolve and HgCl2 will precipitate. NaCl will dissolve and no precipitate will form.

NaCl will dissolve and HgCl2 will precipitate. The solution of HgCl2 is saturated. NaCl is very soluble in water (solubility is over 6 M at 25°C) and will be able to dissolve completely. Because of the common ion effect, as NaCl dissolves, it will decrease the solubility of HgCl2 and cause it to precipitate.

* A sealed container contains NO2, a brownish-red gas, and N2O4, a colorless gas, at equilibrium at 0°C according to the following reaction: . How will the color of the gas mixture change if the container is placed in dry ice and acetone at -78°C? ANSWER INCORRECT Cannot be determined from the information provided. THE CORRECT ANSWER The gas mixture will become lighter in color. The gas mixture's color will not change because it is in equilibrium. YOU WERE SURE AND INCORRECT The gas mixture will become darker in color.

The gas mixture will become lighter in color if the container is placed in dry ice and acetone at -78°C. The system is at equilibrium at 0°C, so ΔG must equal 0. We also know that during the forward reaction 2 moles of gas are converted to 1 mole of gas, which results in a decrease in entropy (ΔS < 0). From this, the sign of ΔH can be determined because ΔG = 0 = ΔH - TΔS. Solving for ΔH gives us: ΔH = TΔS. ΔH must be negative for the forward reaction because ΔS is also negative. Since the forward reaction is exothermic, it will be thermodynamically favored by a decrease in temperature. As temperature falls, the reaction shifts to the right. N2O4 (colorless) increases and NO2 (brownish-red) decreases, leading to an overall lighter color.

The following reaction is allowed to reach equilibrium: Cl2O(g) + 2 OH-(aq) 2 OCl-(aq) + H2O(l) What will be the effect of then doubling the hydroxide concentration? Click here to view the periodic table. ANSWER INCORRECT The reaction quotient will be four times the equilibrium constant, favoring the reverse reaction. YOU WERE SURE AND INCORRECT The reaction quotient will be one-half the equilibrium constant, favoring the forward reaction. The reaction quotient will be twice the equilibrium constant, favoring the forward reaction. THE CORRECT ANSWER The reaction quotient will be one-fourth the equilibrium constant, favoring the forward reaction.

The reaction quotient will be one-fourth the equilibrium constant, favoring the forward reaction. First, eliminate "the reaction quotient will be four times the equilibrium constant, favoring the reverse reaction"; adding a reactant favors the forward reaction. Since the equilibrium constant, K, and the reaction quotient, Q, are inversely proportional to [OH-]2, doubling [OH-] will cause K to decrease by a factor of 22 = 4. Thus, the reaction quotient, Q, will be one-fourth of K, favoring the forward reaction.

Consider the reaction H₂(g)+I₂(g)↔2HI(g) At equilibrium, a 10 L vessel contains 1.5 moles of HI, 0.1 moles of I2 and 0.3 moles of H2. Which direction will the reaction proceed if 2 moles of HI, and 0.5 moles of both I2 and H2 are mixed in a separate 10 L vessel? ANSWER INCORRECT This cannot be determined with the given information. YOU WERE SURE AND INCORRECT The reaction will proceed in reverse. The mixture will be at equilibrium. THE CORRECT ANSWER The reaction will proceed forward.

The reaction will proceed forward All of the information needed to answer this question is given, so the answer choice "this cannot be determined with the given information" can be eliminated. First, Keq can be determined with the equilibrium moles of each compound despite not knowing the total pressure of the vessel. The derivation is shown here for clarity but could be skipped as the expression reduces down to the ratio of moles products to moles reactants at equilibrium raised to the power of their coefficients in the balanced equation: Keq=P(HI)²/ P(H₂) P(I₂) = (XHI x Ptotal)²/(XH₂ x Ptotal) (XI₂ x Ptotal) = (XHI)² / (XH₂) (XI₂) Keq=75 Next, calculate the reaction quotient, Q, for the second vessel in a similar manner: Q=(2)²/(0.5)(0.5) = 16 Since Q < K, the forward reaction is favored.

A reaction between two species is experimentally observed to be second order overall. If the concentration of one of the species doubles, what happens to the reaction rate? ANSWER Cannot be determined The rate is halved The rate quadruples The rate doubles

cannot be determined. There are three possibilities for a two species reaction to produce an overall second order rate law: (1) rate = k[A][B], (2) rate = k[A]2, (3) rate = k[B]2. Let's assume that A is the species doubled. In the first case, doubling A will cause the rate to double. In the second case, doubling A will quadruple the rate. Finally, in the third case, doubling A will cause no change in the rate. Therefore, more information is needed to answer the question properly.

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), ΔH = -22 kcal/mol Increasing the temperature of the reaction at equilibrium will most likely: Click here to view the periodic table. ANSWER INCORRECT decrease the heat of reaction. YOU WERE SURE AND INCORRECT increase the forward reaction. increase the heat of reaction. THE CORRECT ANSWER decrease the forward reaction.

decrease the forward reaction. Since the heat of reaction (ΔH) is not a function of temperature, it will not increase or decrease the heat of reaction. Since the reaction is exothermic, Le Châtelier's principle implies that increasing the temperature will decrease the forward reaction.

As the temperature at which a reaction takes place is increased: Click here to view the periodic table. ANSWER INCORRECT THE CORRECT ANSWER the reaction rate and the rate constant will increase. the reaction rate will remain constant, but the rate constant will increase. the reaction rate and the rate constant will remain constant. YOU WERE SURE AND INCORRECT the reaction rate will increase, but the rate constant will remain constant.

the reaction rate and the rate constant will increase. An increase in the rate constant always corresponds to a faster reaction rate, so the statements "the reaction rate will increase, but the rate constant will remain constant" and "the reaction rate will remain constant, but the rate constant will increase" are eliminated. Increasing the temperature of a reaction always increases the reaction rate, so the statement "the reaction rate and the rate constant will increase" must be correct.