Question -1
C3H6O3(aq) + NaOH(aq) → NaC3H5O3(aq) H2O(l)
A student is studying the reaction between lactic acid, C3H6O3, and sodium hydroxide, NaOH, as represented in the balanced equation above.
(a) The structural formula of lactic acid is shown in the following diagram. Circle the hydrogen atom that most readily participates in the chemical reaction with sodium hydroxide.
(b) The student begins the experiment by dissolving 10.22 g of sodium hydroxide (molar mass 40.00 g / mol) in enough water to produce 500. mL of solution. Calculate the molarity of the sodium hydroxide solution.
The student uses the sodium hydroxide solution from part (b), a buret, a pH meter, and a 100 mL Erlenmeyer flask to titrate a 25.0 mL sample of lactic acid solution. The student’s data are shown in the following graph.
(c) Use the information in the graph to determine the approximate pKa of lactic acid. ____________
(d) The preceding diagram represents the relative amounts of major species in a sample of the solution in the flask at one point during the titration. (Note that water molecules are omitted.)
(i) Draw an X on the preceding titration curve at a point in the titration where the reaction mixture would be represented by this diagram.
(ii) Justify your answer.
(iii) The student repeats the experiment but uses a solution of NaOH(aq) with twice the concentration, as shown in the preceding table. On the following graph, draw the titration curve that would be expected for experiment 2.
(e) In a third experiment, the student investigates the enthalpy of the reaction between lactic acid and sodium hydroxide. The student combines 100.0 mL of a 0.500 M lactic acid solution at 20.0°C with 100.0 mL of a 0.500 M NaOH solution at 20.0°C in a calorimeter. The final temperature of the resulting combined solution is 23.2°C. Assume that the density of each solution before combining is 1.00 g / mL and that the specific heat capacity of the combined solution is 4.2 J / (g · ° C) .
(i) Calculate the quantity of heat produced in the reaction, in J.
(ii) Calculate the molar enthalpy of reaction, in kJ / molrxn. Include the sign in your answer.
(iii) The student claims that if heat is lost from the calorimeter to the surrounding air during the reaction, then the experimental value of the molar enthalpy of reaction will be smaller in magnitude than the actual value. Do you agree or disagree with the student’s claim? Justify your answer
▶️Answer/Explanation
Ans:
(a) For the correct circled atom:
The rightmost hydrogen atom should be circled.
(b) For the correct calculated value:
\(\frac{\left ( 10.22 g \right )\left ( \frac{1 mol}{40.00 g} \right )}{0.500 L} = 0.511 M\)
(c) For the correct pKa :
3.9 (acceptable range: 3.7-4.0)
(d) (i) For the x at current point
The X should be at a point greater than or equal to 3 mL and less than 8 mL.
(ii) For a correct justification:
More acid particles are present than conjugate base particles, meaning that the titration is before the half-equivalence point.
(iii) For a curve showing the correct equivalence point:
The equivalence point should be at 8 mL . See example response below.
For a curve with appropriate initial and final pH with a correct shape:
The drawn curve should begin at the same pH , gradually increase, rise sharply at a volume different than 16 mL , and end at a pH similar to the first curve.
(e) (i) For the correct calculated value:
q = mc∆T = (200.0 g)(4.2 J/(g ⋅ °C))(23.2°C − 20.0°C) = 2700 J
(ii) For the correct calculated value:
qrxn = −qsoln = −2700 J = −2.7 kJ
∆Hrxn = \(\frac{q_{rxn}}{mol}=\frac{-2.7kJ}{(0.100L)(0.500mol/L)}=\) -54kJ/molrxn
(iii) For the correct answer and a valid justification:
Agree. The heat lost from the system would result in a lower final temperature, which results in values of ∆T , qsoln , and ∆H that are smaller than the actual value.
Question-2
A chemical reaction between maleic acid (H2C4H2O4) and sodium bicarbonate (NaHCO3 ) occurs in the presence of water to produce carbon dioxide and sodium maleate (Na2C4H2O4), as represented by the following equation.
H2C4H2O4(aq) + 2NaHCO3(aq) → 2CO2(g) + 2H2O + Na2C4H2O4(aq)
(a) A student combines equal masses of H2C4H2O4(s) chunks and 2NaHCO3(s) chunks with sufficient water at 20.0°C. The student determines that 0.0114 mol of CO2(g) is produced after the reaction goes to completion.
(i) Calculate the number of grams of CO2(g) produced.
(ii) The CO2(g) produced from the reaction at 20.0°C was collected and found to have a pressure of 1.25 atm. Calculate the volume of CO2(g) , in liters.
(b) The student performs a second experiment that is identical to the first except that the student grinds the chunks of H2C4H2O4(s) and NaHCO3(s) into powder before combining the powder with water.
(i) What happens to the surface area of the reactants when the student grinds the chunks into powder?
(ii) The rate-determining step for the overall reaction is the dissolving of the solids. Would the time required for the dissolving of the solids in the second experiment be longer than, shorter than, or the same as the time required in the first experiment? Justify your answer based on the collisions between particles.
(iii) When the reaction is complete, will the volume of CO2(g) at the end of the second experiment be greater than, less than, or equal to the volume at the end of the first experiment? Justify your answer.
The student conducts additional trials of the experiment and produces the following data table.
(c) Based on the student’s data, identify the limiting reactant in trial 3. Justify your answer.
(d) The reaction has a value of ΔS° greater than zero. Using particle-level reasoning, explain why the entropy increases as the reaction progresses.
The student notices that the temperature of the reaction mixture decreases as the reaction takes place and correctly determines that the reaction is endothermic.
(e) The student claims that the reaction is thermodynamically favorable at all temperatures because ΔS°rxn > 0 and the reaction is endothermic. Do you agree or disagree with the student’s claim? Justify your answer.
Next, the student investigates the acid-base behavior of maleic acid. The student notes that maleic acid is a diprotic acid. The two acid dissociation processes that occur are represented by the following equations.
H2C4H2O4 + H2O ↔ HC4H2O4– + H3O+ Ka1=1.5 *10-2
HC4H2O4– + H2O ↔ C4H2O42- + H3O+ Ka2=8.5 *10-7
(f) Calculate the pKa2 value for the HC4H2O4– ion.
(g) A buffer solution with a pH of 7.00 is prepared using C4H2O42- and HC4H2O4–. Calculate the ratio \(\frac{[C_{4}H_{2}O_{4}^{2-}]}{[HC_{4}H_{2}O_{4}^{2-}]}\) in this solution.
▶️Answer/Explanation
Ans:
(a) (i) For the correct calculated value:
0.0114 CO2 * \(\frac{44.01g}{[1 mol]}\) =0.502g CO2
(ii) For the correct calculated value:
PV=nRT
V= \( \frac{nRT}{P} \) = \( \frac{(0.0114mol)( 0.08206 \frac{L.atm}{mol.K})(293K)}{1.25atm} \) = 0.219L
(b) (i) For a correct claim:
The surface area of the solid reactants increases.
(ii) For the correct answer and a valid justification:
Shorter than. The powdered solids have a larger surface area than the solid chunks, thus collisions between water and the surface particles occur more frequently, resulting in a faster rate of dissolution and a shorter amount of time to dissolve the solids.
(iii) For the correct answer and a valid justification:
Equal to. Both experiments begin with the same amount of reactants, so they will produce the same number of moles of CO2 (g) under the same conditions of pressure and temperature; therefore, the final volume will be the same.
(c) For the correct answer and a valid justification:
Accept one of the following:
• NaHCO3 is the limiting reactant because changing the mass of NaHCO3 alters the amount of CO2 produced.
• NaHCO3 is the limiting reactant because the amount present has a smaller theoretical yield of the CO2 product.
\(1.543\\\ g\\\ H_{2}C_{4}H{2}O_{4} \times \frac{1\\\ mol\\\ H_{2}C_{4}H_{2}O_{4}}{116.07\\\ g} \times \frac{2\\\ mol\\\ Co_{2}}{1\\\ mol\\\ H_{2}C_{4}H_{2}O_{4}} = 0.02659\\\ mol\\\ CO_{2}\)
\(1.251\\\ g\\\ NaHCO_{3} \times \frac{1\\\ mol\\\ NaHCO_{3} }{84.01\\\ g} \times \frac{2\\\ mol\\\ CO_{2} }{2\\\ mol\\\ NaHCO_{3}} = 0.02659\\\ mol\\\ CO_{2}\)
(d) For a valid explanation:
The entropy change is positive because the aqueous reactants produce 2 moles of gas particles, according to the balanced chemical equation. Gases are far more dispersed (occupy a greater number of microstates) than condensed phases, so the entropy of the products is greater than that of the reactants.
(e)For the correct answer and a valid justification:
Accept one of the following:
• Disagree. The reaction is endothermic and has a positive entropy change. Thus, the reaction is only thermodynamically favorable at a high enough temperature such that the magnitude of − ∆T S is greater than that of ∆H .
• Disagree. For the reaction to be thermodynamically favorable (∆G < 0) at all temperatures, the reaction must be exothermic (∆H < 0) and have a positive entropy change (∆S > 0)
(f)For the correct calculated value:
pKa2 = −log(8.5 × 10−7 ) = 6.07
(g) For the correct calculated value:
pH= \(pK_{2_{a}} +log \frac{\left [ C_{4}H_{2}O_{4}^{2-} \right ]}{HC_{4}H_{2}O_{4}^{-}}\)
\(\frac{[C_{4}H_{2}O_{4}^{2-}]}{[HC_{4}H_{2}O_{4}^{-}]}={10(^{p^{H}-p^{K_{a2}}})}\) = 10(7.00-6.07) = 8.5
Question-3
Sterling silver is an alloy that is commonly used to make jewelry and consists of 92.5% silver and 7.5% other metals, such as copper, by mass. Over time, the alloy can form a tarnish of Ag2S(s) when it reacts with hydrogen sulfide, as represented by the following equation.
\(2Ag\left ( s \right ) + H_{2}S(g)\to Ag_{2}S(s) + H_{2}(g)\)
(a) What are the oxidation numbers of silver in Ag(s) and Ag2S(s) ?
Ag(s) ___________ Ag2S(s)____________
(b) The following table contains the atomic radii for silver and copper.
Element | Silver(Ag) | Copper(Cu) |
Atomic radius(pm) | 165 | 145 |
(i) Explain why sterling silver is better classified as a substitutional alloy than as an interstitial alloy.
(ii) Using principles of atomic structure and Coulomb’s law, explain why silver has a larger atomic radius than copper does.
The Ag2S tarnish on sterling silver can be removed until only sterling silver remains. A student weighs a tarnished sterling silver sample both before and after removing the Ag2S(s) (molar mass 247.80 g / mol) and records the data in the following table.
(c) Assuming that only Ag2S(s) is removed, calculate the number of moles of silver atoms removed.
Rhodium plating is a process used to protect sterling silver from tarnishing. This involves electroplating (depositing) solid rhodium, Rh(s), onto the surface of the metal from an acidified solution of Rh2(SO4)3 (aq).Oxygen gas is produced during this process.
(d) A table of half-reactions related to the overall reaction is provided.
(i) Write the balanced net ionic equation for plating Rh(s) from the acidified Rh2(SO4)3(aq) solution.
(ii) Calculate the value of E°cell for the reaction in part (d)(i).
(iii) Based on your answer to part (d)(ii), explain why this process requires the use of an external power source.
(e) Calculate the length of time, in seconds, required to plate 2.8 g of Rh(s) onto a piece of sterling silver if 2.0 C/s of current is applied.
▶️Answer/Explanation
Ans:
(a) For the correct answer:
Ag(s) ___0__ Ag2S(s) ___+1___
(b) (i) For a valid explanation:
Silver and copper have similar radii, so the alloy would be substitutional versus interstitial.
(ii) For a valid explanation:
Silver has more occupied electron shells (n = 5) than copper (n = 4); the electrons in the fifth shell experience weaker Coulombic attractions and are farther away from the nucleus.
(c) For the correct calculated mass of Ag2S (may be implicit):
409.21g − 398.94 g = 10.27 g
For the correct calculated moles of Ag :
\(10.27g\times \frac{1 mol Ag_{2}S}{247.80g mol Ag_{2}S} \times \frac{2 mol Ag}{1 mol Ag_{2}S} = 0.08289 mol Ag\)
(d) (i) For the correct balanced equation (state symbols not required):
4 Rh3+ (aq) + 6H2O(l) → 4 Rh(s) + 3O2 (l) + 12 H+ (aq)
(ii) For the correct calculated value, consistent with part (d)(i):
\(E_{cell}^{\circ }\) =+ 0.80 V − 1.23 V = −0.43 V
(iii) For a correct explanation, consistent with part (d)(ii):
\(E_{cell}^{\circ }\) is negative, which means the reaction is not thermodynamically favorable.
(e) For the correct calculated value of moles of electrons (may be implicit):
\(2.8 Rh g\times \frac{1 mol Rh}{102.9g Rh} \times \frac{3 mol e^{-}}{1 mol Rh}=0.082 mol e^{-}\)
For the correct calculated value of time:
\(0.082 mol e^{-} \times \frac{96,485 C}{1 mol e^{-}} \times \frac{1 second}{2.0C}\)= 3900 seconds
Question-4
A student performs an experiment to determine the specific heat capacity of a metal. The student places a cube of the metal in boiling water so its temperature will be 100.0°C. The student then places the metal cube into a calorimeter that contains water and records the highest temperature of the water. A data table and a diagram of the thermometer at the highest temperature are shown.
(a) What should the student report as the highest temperature of the water? _____________
(b) A particle-level representation of water molecules in the calorimeter before and after the metal cube was added is shown. The length of the arrows in the Before diagram represents the speed of the water molecules in the system. In the After diagram, draw an arrow for each molecule to indicate how the speed of each of the molecules changes after the metal cube is added.
(c) Assuming the metal transfers 2940 J of thermal energy to the water, calculate the specific heat of the metal in J/(g·°C) .
(d) In a second experiment, 2940 J of thermal energy is transferred from 98.1 g of aluminum, which has a specific heat capacity of 0.897 J/(g·° C) . Explain how the magnitude of the temperature change of the
aluminum, ΔTAl, compares with the magnitude of the temperature change of the metal in the original experiment.
▶️Answer/Explanation
Ans:
(a) For the correct answer, reported to the correct decimal place:
38.5°C
(b) For a correct drawing:
The “After” drawing should contain arrows that are longer, on average.
(c) For the correct calculated value, consistent with part (a):
q=mcΔT
\(C_{metal}= \frac{q_{metal}}{m_{metal}\bigtriangleup T_{metal}} =\frac{-2940J}{(98.1g)(38.5^{\circ }C-100^{\circ }C)}=0.487\tfrac{J}{g^{\circ }C}\)
(d) For a valid explanation, consistent with part (c):
Accept one of the following:
• The value of ∆TAl will be smaller because Al has a greater specific heat capacity than the metal in the original experiment. Therefore, the same thermal energy transfer applied to the same mass will result in a smaller change in temperature, according to the equation q=mcΔT
• q = mc∆T
\(\left | \bigtriangleup T_{AI}\right | = \left |\frac{q_{AI}}{m_{AI}C_{AI}} \right |=\frac{-2940J}{(98.1g)(38.5^{\circ }C-100^{\circ }C)}= 33.4^{\circ }C\)
\(\left | \bigtriangleup T_{metal}\right | = \left | 38.5^{\circ }C-100.0^{\circ }C\right | = 61.5^{\circ }C\)
Thus, ∆TAl < ∆Tmetal
Question-5
Hydrogen gas and iodine gas react to form hydrogen iodide at an elevated temperature, as represented by the following equation.
H2(g) + I2(g) ⇔ 2HI(g) ΔHrxn = -12.19 kJ/mol Δrxn
(a) Write the expression for the equilibrium constant, Kc, for this reaction.
(b) H2(g) and I2(g) are added to a previously evacuated container and allowed to react.
(i) At a certain time, the value of the reaction quotient, Q, is 0.67. The following particle diagram is an incomplete representation of the system at this time. The diagram shows the relative number of H2(g) and I2(g) molecules, but the HI(g) molecules are not included. Draw the number of HI(g) molecules needed to complete the diagram so that it accurately represents the system.
(ii) A student monitors the number of moles of HI(g) over time. Hypothesize an experimental change that could have been applied to the system in the rigid container at time t to result in the change in the
number of moles of HI(g) shown in the graph. Assume that the student did not add more HI(g) to the system.
(iii) After equilibrium is established, the mixture is transferred to a larger container at constant temperature. As a result, would the number of moles of HI(g) increase, decrease, or remain the same? Justify your answer.
▶️Answer/Explanation
Ans:
(a) For the correct expression:
\( K_{c}= \frac{[HI]^{2}}{[H_{2}][I_{2}]}\)
(b) (i) For the correct drawing consistent with part (a):
(ii) For a valid hypothesis:
Accept one of the following:
• Decreased the temperature.
• Added more H2 and/or I2 to the reaction vessel.
(iii) For the correct answer and a valid justification:
Accept one of the following:
• Remain unchanged. The number of moles in the numerator and denominator of Q (or K ) are equal; changing the volume of the container would not alter the value of Q , which is still equal to K , so the number of moles of HI will remain the same.
• Remain unchanged. The increase in volume will decrease the concentration of reactants and products by an equal proportion. Because there are equal moles of gaseous reactants and products in the balanced chemical equation, there is no shift in the equilibrium position, and the number of moles of HI will remain the same.
Question-6
At elevated temperatures, NO2 undergoes decomposition in the gas phase, forming NO and O2 as represented by the following equation.
2NO2 → 2NO + O2
A scientist measures the change in [NO2] over the first 100. s of the reaction at 546°C. The scientist uses the data collected from the experiment to generate the following two graphs.
Based on these data, the scientist makes the claim that the rate law for the reaction is rate= k[NO2]2 .
(a) Explain how the graphs indicate that the reaction is second order with respect to NO2.
(b) At a certain point in the reaction, the rate of disappearance of NO2 is determined to be 6.52× 10-7 M/s .Determine the rate of appearance, in M/s, of O2 at this same point in the reaction.
(c) NO2 is a molecule that contains an odd number of electrons and can be oxidized to form the NO2+ ion. In NO2, the unpaired electron is presumed to be localized on the nitrogen atom, as shown in the Lewis diagram in the box on the left.
(i) In the box on the right, complete the Lewis diagram for NO2+. Be sure to show all bonding and nonbonding electrons.
(ii) A student makes the claim that the bond angles in NO2 and NO2+ are different from each other. Do you agree or disagree with the student’s claim? Justify your answer.
▶️Answer/Explanation
Ans:
(a) For a correct explanation:
The plot of \(\frac{1}{[NO_{2}]}\) versus time is the most linear, indicating that the reaction is second order with respect to \(NO_{2}./)
(b) For the correct calculated value:
\(6.52 \times 10^{-7}M/s \times \frac{1 mol O_{2}}{2 mol NO_{2}} = 3.26 \times 10^{-7}M/s\)
(c) (i) For the correct Lewis diagram:
(ii) For the correct answer and a valid justification, consistent with part (c)(i):
Accept one of the following:
• Agree. The angle of \(NO_{2}^{+}\) is different from the angle in \(NO_{2}\) because there would no longer be a nonbonding electron on the central atom in \(NO_{2}\) , and the O atoms would spread farther apart, forming a linear structure with a 180° bond angle.
• Agree. The hybridization of N in \(NO_{2}\) is sp2 , which would result in a bond angle of approximately 120° . The hybridization of N in \(NO_{2}^{+}\) is sp , which would result in a bond angle of 180° .
Question-7
A student conducts a chromatography experiment and needs to prepare 100.0 mL of 0.340 M NaCl(aq) to use as the solvent.
(a) Calculate the mass of solid NaCl (molar mass 58.44 g/ mol) needed to prepare the 100.0 mL of 0.340 M NaCl(aq) .
(b) In the following table, briefly list the additional steps necessary to prepare the 100.0 mL of 0.340 M NaCl(aq) solution using only materials selected from the choices given. Assume that all appropriate safety measures are already in place. Not all materials in the list may be needed.
• Solid NaCl • Distilled water • Weighing paper and scoop
• Balance • 100.0 mL volumetric flask • 50.0 mL graduated cylinder
•Pipet •150 mL beakers • Chromatography paper
The student uses the NaCl(aq) solvent to separate a mixture of compounds X and Y in a chromatography experiment. After 30 minutes, the student removes the chromatography paper from the chamber. The results of the experiment are shown.
(c) A second student conducts the same chromatography experiment but removes the chromatography paper from the chamber after 15 minutes instead of 30 minutes. Predict the effect, if any, this would have on the separation distance between compounds X and Y in the new experiment. Explain your reasoning.
▶️Answer/Explanation
Ans:
(a) For the correct calculated value:
\(0.1000L \times \frac{0.340 mol}{1L} \times \frac{58.44g}{1mol} =1.99 g NaCl\)
(b) For a correct description of step 2:
Combine the solid NaCl and some distilled water in a 100.0 mL volumetric flask.
For a correct description of step 4:
Fill the volumetric flask with distilled water to the calibration (100.0 mL) mark.
(c) For the correct prediction and a valid explanation:
It would decrease. The solvent front will not travel as far in the second experiment, so the separation will be smaller.