CHEM 150 CONCEPTS IN CHEMISTRY TEST 2 2017
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CHEM 150
CONCEPTS IN CHEMISTRY
TEST 2
Monday 29th May, 2017 – 6.30 p.m.
QUESTION 1: (20 marks TOTAL)
Calcium phosphate, Ca3(PO4)2 is an acid regulator, used in baking powder and acts as a bread enhancer. It can be prepared by the reaction given below.
3Ca(OH)2(aq) + 2H3 PO4(aq) → Ca3(PO4)2 (s) + 6H2O(aq)
(a) (2 marks) Show that the molar mass of Ca3(PO4)2 is 310.18 g mol-1 .
= 3 × 40.08 + 2 × 30.97 + 8 × 16 = 310.18 g mol-1
(b) (2 marks) Calculate the amount in moles of H3 PO4 present in 25 mL of 0.253
mol L-1 H3 PO4 solution.
= C × V = 0.253 mol L-1 × 0.025 L = 6.325 × 10-3 mol
(c) (2 marks) Calculate the moles of Ca(OH)2 required to completely react with the H3 PO4 in the solution in (b) above.
6.325 10一3 mol 3
n(Ca(OH)2 = 3 = 9.4875 10一 mol
2
(d) (3 marks) Using your answer to (c) above calculate the mass of Ca3(PO4)2 produced in this reaction.
n(Ca3(PO4)2) = 9.4875 × 10-3 / 3 =3.1625 × 10-3 mol
m(Ca3(PO4)2) = 3.1625 × 10-3 mol × 310.18 g mol-1 = 0.98 g
(e) (2 marks) A technician prepared 2.0 × 10-2 mol L-1 FeSO4 stock solution. For a
spectroscopic analysis a student needs to dilute the stock solution to prepare 100 mL of 5.0 × 10-3 mol L-1 FeSO4 solution. Calculate the volume of stock solution that is required to prepare the diluted solution.
n(FeSO4) required = 5.0 × 10-3 mol L-1 × 0.1 L = 5 × 10-4 mol V(FeSO4) = = 5 × 10-4 mol / 2.0 × 10-2 mol L-1 = 0.025 L = 25.0 mL
(f) Potassium bitartrate, K2C4 H4O6 is commonly known as cream of tartar. This is
produced when tartaric acid, C4 H6O6 is titrated with NaOH as shown below. C4 H6O6 (aq) + 2KOH(aq) → K2C4 H4O6(aq) + 2H2O(l)
In a titration 20.0 mL of tartaric acid required 21.35 mL of 0.152 mol L-1 KOH for complete neutralisation.
(i) (1 mark) Calculate the amount in moles of KOH used in this titration.
= 0.152 mol L_1 × 21.35 × 10-3 L = 3..2452 × 10-3 mol
(ii) (1 mark) Calculate the amount in moles of tartaric acid, C4 H6O6 in 20.0
mL sample.
3.245210一3 mol 3
n(C4 H6O6 ) = 2 1 = 1.6226 10一 mol
(iii) (1 mark) Calculate the concentration of the tartaric acid C4 H6O6
solution.
C(C4 H6O6 ) = 0一3(一)3L(mol) = 0.0811 mol L一1
(iv) (1 mark) The molar mass of tartaric acid 150.087 g mol-1 . Express the
concentration of tartaric acid you calculated in (iii) above in g L-1 .
Concentration = 0.0811 mol L-1 × 150.087 g mol-1 = 12.17 gL-1
(g) Nitrogen gas can be prepared by passing NH3 gas over solid CuO at high
temperature as shown by the following equation.
2NH3(g) + 3CuO → N2(g) + 3Cu(s) + 3H2O(g)
In one experiment 18.1 g NH3 is allowed to react with 90.4 g CuO.
(i) (3 marks) Determine the limiting reactant in this preparation. Show all your working. M(NH3) = 17.03 g mol-1 , M(CuO) = 79.55 g mol-1
18.1g 90.4 g
n(NH3 ) = -1 = 1.063 mol , n(CuO) = -1 = 1.136 mol
17.03 g mol 79.55 g mol
1.063 mol
2
have only 1.136 mol of CuO. So CuO is the limiting reactant.
(ii) (2 marks) Calculate the mass of N2 produced in this experiment. M(N2)
28.02 g mol-1
n(CuO) 1.136 mol
n(N2 ) = = = 0.3786 mol
2 3
M(N2) = 0.3786 mol × 28.02 g mol-1 = 10.6 g
QUESTION 2 (20 marks)
(a) (1 mark) Give an example for an intensive property.
Temperature, density
(b) |
Diborane, B2 H6 mixes with air, easily forming explosive mixtures. It ignites at room temperature as shown in the equation below. B2 H6 (g) + 3O2(g) → B2O3 (s) + 3H2O(l) H° = -2148 kJ |
(i) (2 marks) If the above reaction produced H2O(g) instead of H2O(l) would the enthalpy change be same, more negative or less negative?
Justify your answer
Less negative as some energy will be used to convert H2O(l) to H2O(g). OR vaporisation is an endothermic process.
(ii) (3 marks) Calculate the enthalpy of formation of B2 H6 using the equation
in (b) above and the information given below .
Substance |
fH° |
B2O3(s) |
-1255 |
H2O(l) |
-286 |
(c) (3 marks) Calculate the enthalpy change for the following reaction using the reaction enthalpies given.
CaCO3(s) → CaO(s) + CO2(g)
CaO(s) + 2HCl(aq) → CaCl2(aq) + H2O(l)
H° = ?
H° = - 386 kJ
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) H° = - 573 kJ
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) CaCl2(aq) + H2O(l) → CaO(s) + 2HCl(aq) |
H° = - 573 kJ H° = 386 kJ |
|
CaCO3(s) → CaO(s) + CO2(g) |
H° = -573 – 386 kJ = -187 kJ |
(c) Phosphine, PH3 is used in the semiconductor industry to introduce phosphorus
into silicon crystals. It is prepared by the reaction shown below. P4(g) + 6H2(g) → 4PH3(g)
In one preparation the consumption of H2 gas was measured. It is found that the concentration of H2 dropped from 1.75 mol L-1 to 0.43 mol L-1 within 30 seconds.
(i) (1 mark) Calculate the rate of consumption of H2 in this preparation.
= 一 (0.43 一 一1 = 0.044 mol L一1 s一1
(ii) (2 marks) Calculate the rate of production of PH3 .
= 4 = 0.0.44 m6(o)l L一1 s一1 4 = 0.029 mol L一1 s一1
(iii) (2 mark) What is the overall rate of this reaction.
0.044 mol L-1 s-1/ 6 or 0.029 mol L-1 s-1 / 4 = 7.33 × 10-3 mol L-1 s-1
(iv) The rate law for the above reaction is determined to be
Rate = k[P4][H2]2
a. (1 mark) What happens to the rate of reaction if the concentration of P4 is tripled?
Rate will increase by a factor of 3 / tripled |
b. (2 marks) What happens to the rate of reaction if the concentration of H2 is doubled? Justify your answer.
Rate will increase by a factor of 4 Rate [ ]2 = 22 = 4 |
c. (3 marks) In an experiment when the concentrations of P4 and H2 are 0.012 mol L-1 and 0.156 mol L-1 the rate of reaction is found to be 1.26 mol L-1s-1 . Calculate the rate constant for this reaction. Give the units of the rate constant.
Rate = k[P4][H2]2
1.26 mol L-1s-1 = k × 0.012 mol L-1 × [0.156 mol L-1]2
1.26 mol L-1 s-1
0.012 mol L [0. 156 mol L ]
QUESTION 3 (10 marks)
(a) (1 mark) Write an expression for the equilibrium constant for the following
reaction.
Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(g)
Kc = [H2O]3 / [H2]3 |
(b) (1 mark) For a specific reaction, which of the following statements can be made
about K, the equilibrium constant? Circle the right choice.
(i) It always remains the same at different reaction conditions.
(ii) It increases if the concentration of one of the products is increased. (iii) It changes with changes in the temperature.
(iv) It increases if the concentration of one of the reactants is increased.
(v) It may be changed by the addition of a catalyst.
Justify your answer
Answer: (iii). For an endothermic reaction, an increase in temperature will shift the position of equilibrium toward the products, which will increase K. For exothermic reaction, an increase in temperature will shift the position of equilibrium toward the reactants, which will decrease K. |
(c) (3 marks) On analysis, an equilibrium mixture for the reaction 2H2S(g) 2H2(g) + S2(g) was found to contain 1.0 mol H2S, 4.0 mol H2 , and 0.80 mol S2 in a 4.0 L vessel. Calculate the equilibrium constant (K) for this reaction.
[H2S] = 1.0mol/4.0L = 0.25 mol/L, [H2] = 4.0mol/4.0L = 1.0 mol/L [S2] = 0.8mol/4.0L = 0.2 mol/L K=[H2]2 [S2]/ [H2S]2 = (1.0)2(0.2)/(0.25)2 = 0.2/0.0625 = 3.2 |
(d) (2 marks) Hydrogen gas and chlorine gas in the presence of light react explosively to form hydrogen chloride:
H2(g) + Cl2(g) |
|
2HCl(g) |
The reaction is strongly exothermic. Would an increase in temperature for the system tend to favour or disfavour the production of hydrogen chloride?
As the reaction is exothermic, heat is effectively a product. An increase in temperature would basically fight against of forward reaction as it tries to release energy, and would disfavour production of HCl |
(e) (3 marks) For parts (1) to (3), write the number of the best answer on the line
provided. Each part is worth 1 mark.
1. If the reaction quotient (Q) has a smaller value than the related equilibrium constant (K)
(i) the reaction is at equilibrium.
(ii) the reaction is not at equilibrium, and will make more products at the
expense of reactants.
(iii) the reaction is not at equilibrium, and will make more reactants at the
expense of products.
2. Which of the following occurs when reactants are added to a chemical reaction in solution or the gas phase at equilibrium?
(i) Q increases, so the equilibrium shifts to produce more products. (ii) Q increases, so the equilibrium shifts to produce more reactants. (iii) Q decreases, so the equilibrium shifts to produce more products.
(iv) Q decreases, so the equilibrium shifts to produce more reactants.
(v) Q is unchanged by the addition of reactants
3. Which of the following are equal for a chemical system at equilibrium?
(i) the concentrations of reactant and products are equal
(ii) the rate constants for the forward and reverse reactions are equal
(iii) the time that a particular atom or molecule spends as a reactant and product
are equal
(iv) the rate of the forward and reverse reactions are equal
(v) all of the above are equal
2023-06-02