Question
\(CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)\)
When heated, calcium carbonate decomposes according to the equation above. In a study of the decomposition of calcium carbonate, a student added a 50.0 g sample of powdered \(CaCO_{3}\)(s) to a 1.00 L rigid container. The student sealed the container, pumped out all the gases, then heated the container in an oven at 1100 K. As the container was heated, the total pressure of the \(CO_2\)(g) in the container was measured over time. The data are plotted in the graph below.
The student repeated the experiment, but this time the student used a 100.0 g sample of powdered \(CaCO_3\)(s). In this experiment, the final pressure in the container was 1.04 atm, which was the same final pressure as in the first experiment.
(a) Calculate the number of moles of \(CO_{2}(g)\) present in the container after 20 minutes of heating.
(b) The student claimed that the final pressure in the container in each experiment became constant because all of the\( CaCO_{3}\)(s) had decomposed. Based on the data in the experiments, do you agree with this claim? Explain.
(c) After 20 minutes some \(CO_2\)(g) was injected into the container, initially raising the pressure to 1.5 atm. Would the final pressure inside the container be less than, greater than, or equal to 1.04 atm? Explain your reasoning.
(d) Are there sufficient data obtained in the experiments to determine the value of the equilibrium constant, K,for the decomposition of \(CaCO_3\)(s) at 1100 K? Justify your answer.
▶️Answer/Explanation
(a) \(PV=nRT\)
\(\frac{PV}{RT}=n=\frac{(1.04 atm)(1.00 L)}{(0.0821\frac{L atm}{mol K})(1100 K)}== 0.0115 mol CO_2\)
(b) Do not agree with claim
Explanation I: In experiment 1, the moles of \(CaCO_3 = 50.0 g/100.09 g/mol = 0.500 mol CaCO_3\). If the reaction had gone to completion, 0.500 mol of \(CO_2\) would have been produced. From part (a) only 0.0115 mol was produced. Hence, the student’s claim was false.
Explanation II: The two different experiments (one with 50.0 g of \(CaCO_3\) and one with 100.0 g of \(CaCO_3\)) reached the same constant, final pressure of 1.04 atm. Since increasing the amount of reactant did not produce more product, there is no way that all of the \(CaCO\), reacted. Instead, an equilibrium condition has been achieved and there must be some solid \(CaCO_3\) in the container.
(c) The final pressure would be equal to 1.04 atm. Equilibrium was reached in both experiments; the equilibrium pressure at this temperature is 1.04 atm. As the reaction shifts toward the reactant, the amount of \(CO_2\)(g) in the container will decrease until the pressure returns to 1.04 atm.
(d) Yes. For the equilibrium reaction represented by the chemical equation in this problem, at a given temperature the equilibrium pressure of \(CO_2\) determines the equilibrium constant. Since the measured pressure of \(CO_2\) is also the equilibrium pressure of \(CO_2\), \(K_p= P_{co_2} =1.04\).
Note: If the response in part (b) indicates “yes”, that all of the \(CaCO_3\)(s) had decomposed, then the point can be earned by stating that the system did not reach equilibrium in either experiment and hence the value of \(K_p\), cannot be calculated from the data.
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
\(NH_3(aq) + H_2O(I) \leftrightarrow NH_4^+(aq) + OH^-(aq)\)
In aqueous solution, ammonia reacts as represented above. In 0.0180 M \(NH_3(aq)\) at \(25^o\)C, the hydroxide ion concentration, [\(OH^ -\)] , is \(5.60 x 10^{-4}\) M. In answering the following, assume that temperature is constant at 25°C and that volumes are additive.
(a) Write the equilibrium-constant expression for the reaction represented above.
(b) Determine the pH of 0.0180 M \(NH_3(aq)\).
(c) Determine the value of the base ionization constant, \(K_b\) , for \(NH_3(aq)\).
(d) Determine the percent ionization of \(NH_3\) in 0.0180 M \(NH_3(aq)\).
(e) In an experiment, a 20.0 mL sample of 0.0180 M \(NH_3(aq)\) was placed in a flask and titrated to the equivalence point and beyond using 0.0120 M HCl(aq).
(i) Determine the volume of 0.0120 M HCl(aq) that was added to reach the equivalence point.
(ii) Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120 M HCl(aq) was added.
(iii) Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120 M HCl(aq) was added.
▶️Answer/Explanation
Answer:
(a) \(K = \frac{[NH_4^+][OH^-]}{[NH_3]}\)
(c) \(K_b =\frac{(5.60 \times 10^{-4})^2}{0.0180-5.60 \times 10^{-4}}= 1.74 \times 10^{-5}\) (or\(1.80 \times 10^{-5}\))
Note : \(1^{st}\) point for \([NH_4^+]=[OH^-]=5.60 \times 10^{-4};2^{nd}\) point for correct answer
(d) % ionization \(=\frac{5.60 \times 10^{-4}}{0.0180} \times 100 % = 6.11 %\) (or 0.311)
(e) \(NH_3 + H^+ \rightarrow NH_4^+\)
(i) mol \(NH_3 = 0.0180\) \( M \times 0.0200 L = 3.60 \times 10^^{-4}\) mol = mol \(H^+\) needed
vol HCl solution = \(\frac{3.60 \times 10^{-4} mol}{0.0120M}\) = 0.0300 L = 30.0 mL
(ii) mol \(H^+\) added = mol \(0.0120 M \times 0.0150 L = 1.80 \times 10^{-4} mol H^+ added \)
\(= 1.80 \times 10^{-4} mol NH_4^+produced\)
\([NH_4^+]=\frac{1.80 \times 10^{-4}mol}{0.0350 L}=0.00514 M = [NH_3]\)
Note : Point earned for \(1.80 \times 10^{-4}mol\), or \(0.00514 M[NH_3]or[NH_4^+]\),
or statement “halfway to equivalence point”.
\(K_b=1.80 \times 10^{-5} = \frac{[NH_4^+][OH^-]}{[NH_3]}=[ON^-] \Rightarrow pOH = 4.745 \Rightarrow pH = 9.255\)
\((=1.74 \times 10^{-5})(=4.759)(=9.241)\)
(iii) 10.0 mL past equivalence point
\(0.0100 L \times 0.0120 M = 1.20 \times 10^{-4} mol H^+ in 60.0mL\)
\([H^+]=\frac{0.000120mol}{0.0600 L}=0.00200M\)
\(pH = – log(2.00 \times 10^{-3})=2.700\)