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These chapters come from an earlier course, and this one relies on them from the first weeks.
202-SN1-RE · Measurement, significant figures and uncertainty
The figure shows five pieces of glassware with the tolerance stated by the manufacturer.
c) An aliquot is to be taken for a dilution: the volume taken will go directly into the calculation of the final concentration. Which of the five instruments should be chosen?
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Answer
c)The 25.00 mL volumetric pipette
c) The volumetric pipette. It is the only instrument on the list designed to DELIVER a single, fixed volume with high precision, which is exactly what an aliquot requires. The volumetric flask, on the other hand, is used to CONTAIN a volume, not to deliver one: the diluted solution is prepared in it, nothing is withdrawn from it.
202-SN1-RE · Stoichiometry, limiting reactant and percent yield
Ammonia is synthesized according to N2+3H2→2NH3. 28.0 g of dinitrogen N2 are reacted with 5.00 g of dihydrogen H2. Molar masses (g/mol): N2 28.02; H2 2.02; NH3 17.04.
The diagram below shows the synthesis.
b) Determine the limiting reactant by comparing the ratios coefficientn.
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Answer
b)0.825<1.00: dihydrogen is limiting
b) For N2: 11.00=1.00. For H2: 32.48=0.825. The smallest ratio is that of dihydrogen, so H2 is the limiting reactant: it runs out before the dinitrogen. Once the limiting reactant is used up, adding more excess reactant produces nothing more: the amount of product levels off on a plateau.
In solution, an amount of substance reads as a product C×V. All the stoichiometry of the previous chapter then applies unchanged, provided you never confuse the volume of solution with the volume of solute.
Molar mass of AgCl: 143.32 g/mol. The diagram below shows the titration of b) and the precipitation of c).
a) How would you prepare 500 mL of hydrochloric acid at 0.150 mol/L from a commercial solution at 12.0 mol/L? Describe the procedure.
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Answer
a)Withdraw 6.25 mL and make up to 500 mL in a flask that already contains water
a) Dilution conserves the amount of substance: C1V1=C2V2, so V1=C1C2V2=12.00.150×0.500=6.25×10−3 L, or 6.25 mL. Procedure: draw 6.25 mL of the commercial solution with a pipette, transfer it to a 500 mL volumetric flask that ALREADY contains some distilled water, then fill up to the calibration mark and mix. Never put the acid into an empty flask that you then fill with water.
A system exchanges energy with its surroundings in only two ways: as heat and as work. The first law states that the sum of the two is a state function, the internal energy.
Reminder of units: 1 kPa·L = 1 J.
The diagram below shows the expansion in b).
b) A gas expands from 2.00 L to 5.00 L against a constant external pressure of 101.3 kPa, absorbing 500 J of heat. Calculate the work, then the change in internal energy.
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Answer
b)w=−304 J and ΔU=+196 J
b) ΔV=5.00−2.00=+3.00 L. The work is w=−PΔV=−101.3×3.00=−304 kPa·L, that is −304 J. The sign is negative because the gas PUSHES the atmosphere: it does work. Then ΔU=q+w=500+(−304)=+196 J.
Chapter 1 · Measurement, uncertainty and graphical analysis
We now have the calibration line A=aC+b with a=(0.191±0.008) L/(mg⋅cm) and b=0.000, established for C between 0.500 and 2.500 mg/L. A river water sample is treated with the same reagent: 5.00 mL is taken with the pipette (±0.015 mL) and made up to 100.00 mL (±0.08 mL), and the cuvette gives A=0.305±0.008.
The diagram below shows how the sample is prepared.
b) Calculate its relative and then absolute uncertainty, and write the publishable result.
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Answer
b)6.8%, that is C=(1.6±0.1) mg/L; the slope dominates
b) The quotient aA makes the RELATIVE uncertainties add: 0.3050.008×100=2.6% for the absorbance, 4.2% for the slope, total 6.8%. Absolute uncertainty: 0.068×1.596=0.11 mg/L, rounded to one significant figure, so 0.1 mg/L. Publishable result: C=(1.6±0.1) mg/L. Two digits, not four. Note which of the two contributions dominates: the SLOPE, for nearly two thirds. Improving the absorbance reading would gain almost nothing as long as the calibration curve stays what it is, which is the opposite of intuition, because the reading is what we have just taken and the curve dates from an hour ago.
a) Determine the order of the reaction with respect to A, then with respect to B, spelling out the comparison you use.
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Answer
a)Order 2 in A, order 1 in B
a) From experiment 1 to experiment 2: [A] doubles with [B] constant, and v is multiplied by 2.08.0=4=22: order 2 in A. From experiment 1 to experiment 3: [B] doubles with [A] constant, and v is multiplied by 2: order 1 in B. Always change only ONE concentration at a time: that is what isolates each exponent.
At 430 °C, the reaction H2(g)+I2(g)⇌2HI(g) has a constant Kc=54.3.
The diagram below shows the mixture in c).
c) A mixture with [H2]=[I2]=[HI]=0.10 mol/L is prepared at 430 °C. Calculate the reaction quotient and predict the direction of change.
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Answer
c)Q=1.0<Kc=54.3, change in the forward direction
c) Q=(0.10)(0.10)(0.10)2=1.0. Since Q=1.0<Kc=54.3, the numerator has to grow: the system moves in the FORWARD direction, making HI by consuming H2 and I2, until Q reaches 54.3.
d) Is the reaction HF+CN−⇌F−+HCN favoured to the right or to the left? Justify with the Ka values, without calculating a constant.
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Answer
d)To the right: ratio of the Ka values close to 106, strongly favoured
d) To the RIGHT. The forward reaction sets the acid HF (Ka=6.6×10−4) against the acid HCN (Ka=6.2×10−10): the reaction proceeds from the stronger acid to the weaker acid, and the proton always goes down to the hungrier base (here CN−). The ratio of the two Ka values, about 106, would in fact give the equilibrium constant of the reaction directly: strongly favoured.
A 20.00 mL sample of 0.150 M ammonia solution is titrated with 0.100 M hydrochloric acid. Take Kb(NH3)=1.8×10−5.
The diagram below shows the setup.
a) Write the titration equation and calculate the equivalence volume.
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Answer
a)NH3+H3O+→NH4++H2O, Veq=30.00 mL
a) NH3+H3O+→NH4++H2O. Equivalence: n(NH3)=n(HCl), that is 0.150×20.00=0.100×Veq, so Veq=30.00 mL. Be careful, the concentrations are DIFFERENT here, and the equivalence volume has no reason to equal the initial volume: this is a classic error for anyone who memorized the previous exercise instead of the method.
In the laboratory, one litre of saturated water is found to dissolve 2.40×10−3 g of barium sulfate BaSO4 (M=233.40 g/mol) and 2.16×10−2 g of silver chromate Ag2CrO4 (M=331.74 g/mol).
The diagram below shows the two saturated solutions.
b) Calculate the Ksp of Ag2CrO4. Mind the stoichiometry.
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Answer
b)s=6.51×10−5 mol/L and Ksp=4s3=1.10×10−12
b) s=331.742.16×10−2=6.51×10−5 mol/L. Here Ag2CrO4(s)⇌2Ag++CrO42−: each formula unit dissolved releases TWO silver ions, so [Ag+]=2s and [CrO42−]=s. Ksp=(2s)2(s)=4s3=4(6.51×10−5)3=1.10×10−12. Forgetting the coefficient 2 (in the concentration AND in the exponent) is the most common error of the chapter: here it throws the result off by a factor of 4.
The oxidation number is a bookkeeping convention: the electrons of each bond are assigned to the more electronegative atom. Two exceptions trip up half the papers, and they come back every term.
b) Determine that of oxygen in H2O2 and that of hydrogen in NaH. Name the exception in each case.
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Answer
b)n(O)=−1 in H2O2 (peroxide) and n(H)=−1 in NaH (hydride)
b) In H2O2: the two H are +1, the molecule is neutral, so 2(+1)+2n(O)=0 and n(O)=−1. This is the PEROXIDE exception, where two oxygens are bonded to each other. In NaH: Na is an alkali metal, it is +1, so n(H)=−1. This is the metal HYDRIDE exception, where hydrogen is bonded to an element LESS electronegative than itself.
ΔG=ΔH−TΔS combines the two driving forces of a change: the energy released and the disorder produced. Their relative importance depends on temperature, which gives four possible behaviours.
For CaCO3(s)→CaO(s)+CO2(g): ΔHr∘=+179.2 kJ/mol and ΔSr∘=+160.6 J/(mol·K).
a) Draw up the table of the four cases according to the signs of ΔH and ΔS, stating for each the temperature range in which the reaction is spontaneous.
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Answer
a)Always, never, at low temperature, at high temperature, according to the four combinations of signs
a) Case 1, ΔH<0 and ΔS>0: both terms favour the reaction, ΔG<0 at EVERY temperature, the reaction is always spontaneous. Case 2, ΔH>0 and ΔS<0: both oppose it, ΔG>0 always, NEVER spontaneous. Case 3, ΔH<0 and ΔS<0: enthalpy favours, entropy opposes; since the entropy term is multiplied by T, it wins as T grows: spontaneous at LOW temperature. Case 4, ΔH>0 and ΔS>0: the opposite situation, spontaneous at HIGH temperature.
A dissolution is spontaneous if ΔGsol=ΔHsol−TΔSsol is NEGATIVE. Enthalpy alone therefore decides nothing.
Data at 25 °C. Ammonium nitrate: ΔHsol=+26 kJ/mol and ΔSsol=+108 J/(mol·K). Oxygen dissolved in water: ΔHsol=−12.5 kJ/mol and ΔSsol=−95 J/(mol·K).
a) Calculate ΔGsol of ammonium nitrate at 25 °C, in kJ/mol. State your conclusion.
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Answer
a)ΔGsol=−6.2 kJ/mol: spontaneous despite the endothermicity
a) ΔGsol=26000−298×108=26000−32184=−6184 J/mol, that is −6.2 kJ/mol. The dissolution is therefore SPONTANEOUS even though it cools the solution: entropy pays the bill. Watch out for the unit trap, which costs a mark on every exam: the enthalpy is in kilojoules, the entropy in joules per kelvin. Without converting, you get 26−32184=−32158 kJ/mol, an absurd value that no order-of-magnitude check would let through.
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