Question.
\(H_{3}BO_{3}(aq)+4HF(g)\rightarrow H_{3}O^{+}(aq)+BF_{4}^{-}(aq)+2H_{2}O(l)\)
Tetrafluoroboric acid is a strong acid with the formula\( HBF_4\). The acid can be prepared by reacting the weak acid \(H_3BO_3\) (molar mass 61.83 g/mol) with HF according to the equation above.
(a) To prepare a solution of \(BF_4^{-}\)(aq), HF(g) is bubbled into a solution containing 50.0 g of\( H_3BO_3\) in a 1 L reaction vessel.
(i) Calculate the maximum number of moles of\( BF_4^{-}\)(aq) that can be produced.
(ii) Calculate the number of liters of HF(g), measured at 273 K and 1.00 atm, that will be consumed if all the\( H_3BO_3\) reacts.
(iii) Will the pH of the solution increase, decrease, or remain the same during the course of the reaction? Justify your answer. In another experiment, a 0.150 M \(BF_4^-\)(aq) solution is prepared by dissolving \(NaBF_4\)(s) in distilled water. The\( BF_{4}^-\)(aq) ions in the solution slowly react with H2O(l) in the reversible reaction represented below.
\(BF_4^-(aq)+H_{2}O(l)\rightleftharpoons BF_{3}OH^{-}(aq)+HF(aq)\)
The concentration of HF is monitored over time, as shown in the graph below.
[HF] reaches a constant value of 0.0174 M when the reaction reaches equilibrium. For the forward reaction, the rate law is rate = kf \([BF_4^{-}\)]. The value of the rate constant kf was experimentally determined to be \(9.00\times 10^{-4}min^{-1}\).
(b) Calculate the rate of the forward reaction after 600. minutes. Include units with your answer. The rate law for the reverse reaction is rate = \(kr [BF_3OH^-]\)[HF].
(c) A student claims that the initial rate of the reverse reaction is equal to zero. Do you agree or disagree with this claim? Justify your answer in terms of the rate law for the reverse reaction.
(d) At equilibrium the forward and reverse reaction rates are equal. Calculate the value of the rate constant for the reverse reaction.
\(H_{3}BO_{3}(aq)+4HF(g)\rightarrow H_{3}O^{+}(aq)+BF_{4}^{-}(aq)+2H_{2}O(l)\)
Tetrafluoroboric acid is a strong acid with the formula \(HBF_4\). The acid can be prepared by reacting the weak acid \(H_3BO_3\) (molar mass 61.83 g/mol) with HF according to the equation above.
▶️Answer/Explanation
a(i) \(50.0gH_{3}BO_{3}\times \frac{1molH_{3}BO_{3}}{61.83g}\times \frac{1molBF_{4}^{-}}{1molH_{3}BO_{3}}\)
a(ii)\( 0.809 mol \times 4 = 3.24 mol\)
PV= nRT
\(V=\frac{nRT}{p}=\frac{(3.24mol)(0.08206Latm mol^{-1}K^{-1})(273K)}{1.00atm}\)
=72.6L
OR
\(0.809 mol \times 4 = 3.24 mol\)
\(\frac{22.4L}{1mol}\times 3.24mol=72.6 L\)
a(iii) As the reaction proceeds,\( H_3O^{+}\) is produced, so the pH
will decrease.
(b) \([BF_4^{-}] = 0.150 M – 0.0174 M = 0.133 M\)
\(rate = (9.00\times 10^{-4}min^{-1})(0.133M)=1.20\times 10^{-4}M min^{-1}\)
(c) Agree. The initial concentration of each product is zero, so the initial rate of the reverse reaction is zero.
(d ) 
Question
Answer the following questions relating to HCl,\(CH_3Cl,\) and \(CH_3\)Br.
(a) HCl(g) can be prepared by the reaction of concentrated \(H_ 2SO_4\)(aq) with NaCl(s), as represented by the following equation.
\(H_{2}SO_{4}(aq)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq)\)
(i) A student claims that the reaction is a redox reaction. Is the student correct? Justify your answer.
(ii) Calculate the mass, in grams, of NaCl(s) needed to react with excess \(H_2SO_4\)(aq) to produce 3.00 g of HCl(g). Assume that the reaction goes to completion.HCl(g) can react with methanol vapor,\( CH_3OH(g)\), to produce \(CH_3Cl\)(g), as represented by the following equation.
\(CH_{3}OH(g)+HCl(g)\rightleftharpoons CH_{3}Cl(g)+H_{2}O(g) K_{P}=4.7\times 10^{3}\) At400K
(b) \(CH _3OH(g)\) and HCl(g) are combined in a 10.00 L sealed reaction vessel and allowed to reach equilibrium at 400 K. The initial partial pressure of\( CH_3OH(g)\) in the vessel is 0.250 atm and that of HCl(g) is 0.600 atm.
(i) Does the total pressure in the vessel increase, decrease, or remain the same as equilibrium is approached? Justify your answer in terms of the reaction stoichiometry.
(ii) Considering the value of Kp , calculate the final partial pressure of HCl(g) after the system inside the vessel reaches equilibrium at 400 K.
(iii) The student claims that the final partial pressure of \(CH_3OH(g) \)at equilibrium is very small but not exactly zero. Do you agree or disagree with the student’s claim? Justify your answer.
(c) The table below shows some data for the compounds \(CH _3Cl \)and \(CH_3Br\).
(i) Identify all the types of intermolecular forces that exist among molecules in \(CH_3Cl(l)\).
(ii) In terms of intermolecular forces, explain why the boiling point of CH3Br(l) is greater than that of \(CH_3Cl(l)\).
(d) A 2.00 mL sealed glass vial containing a 1.00 g sample of \(CH_3Cl(l)\) is stored in a freezer at 233 K.
(i) Calculate the pressure in the vial at 298 K assuming that all the \(CH_3Cl(l)\)vaporizes.
(ii) Explain why it would be unsafe to remove the vial from the freezer and leave it on a lab bench at 298 K.
▶️Answer/Explanation
a(i) No, the student is not correct. None of the oxidation numbers of the elements change (H = +1, S = +6, O = -2, Na = +1, Cl = -1).
a(ii) 
(b)(i) The pressure will remain the same. The reaction stoichiometry shows that two moles of gaseous reactants produce two moles of gaseous products. Because the number of moles of gas does not change, the pressure does not change.
b(ii)The value of\( K_p\) is large, so the reaction will proceed to the right until the limiting reactant is essentially used up. Thus practically all of the \(CH_3OH(g)\) will react and the final pressure of HCl(g) is 0.600 – 0.250 = 0.350 atm.
OR
The final pressure of HCl(g) is 0.350 atm at equilibrium.
b(iii) Agree. The large value of \(K_p\) means that the partial pressure of the limiting reactant at equilibrium will be extremely small, but some \(CH_3OH \)molecules must exist for the system to be in dynamic equilibrium.
The partial pressure of \(CH_3OH(g)\) is very small but is not zero.
c(i) London dispersion forces and dipole-dipole forces
c(ii) The electron cloud in \(CH_3Br\) is larger and more polarizable than that of \(CH_3Cl\). As a result the London dispersion forces are stronger in \(CH_3Br\) compared to those in \(CH_3Cl\)and consequently the boiling point of\( CH_3\)Br is higher than that of \(CH_3Cl\).
d(i)
d(ii) At room temperature the liquid will vaporize. Consequently the glass vial may not be strong enough to withstand the increase in pressure.
Question
Consider the flasks in the following diagram.
(a) What are the final partial pressures of He and Ar after the stopcock between the two flasks is
opened? (Assume that the final volume is 4.00 L.)
(b) What is the total pressure (in torr)?
▶️Answer/Explanation
Ans:
