Paper 1 2018 Flashcards
(31 cards)
A sample of boron contains the isotopes 10-B and 11-B. The relative atomic mass of the boron sample is 10.8. What is the percentage of 11-B atoms in the sample of boron?
80%
In the compound [ICl2] + [SbCl6]–, the oxidation number of chlorine is –1. What are the oxidation numbers of I and Sb in the compound?
- I = +3
- Sb = +5
What is the number of hydrogen atoms in 0.125 mol of C2H5OH?
4.515 x 10^23
A student titrates a standard solution of barium hydroxide, Ba(OH)2, with nitric acid, HNO3.
25.00 cm3 of 0.0450 mol dm–3 Ba(OH)2 are needed to neutralise 23.35 cm3 of HNO3(aq).
What is the concentration, in mol dm–3, of the nitric acid?
0.0964
Which statement best explains why nitrogen has a larger first ionisation energy than oxygen?
A: N atoms have less repulsion between p-orbital electrons than O atoms.
B: N atoms have a smaller nuclear charge than O atoms.
C: N atoms lose an electron from the 2s subshell, while O atoms lose an electron from the 2p subshell.
D: N atoms have an odd number of electrons, while O atoms have an even number.
A
In the Periodic Table, element X is in Group 2 and element Y is in Group 15 (5). What is the likely formula of an ionic compound of X and Y?
X3Y2
Which statement about ammonium carbonate is not correct?
A: It reacts with Ba(NO3)2(aq) to form a white precipitate.
B: It effervesces with dilute nitric acid.
C: It releases an alkaline gas with warm NaOH(aq).
D: It has the formula NH4CO3.
D
The reversible reaction of sulfur dioxide and oxygen to form sulfur trioxide is shown below.
2SO2(g) + O2(g) 2SO3(g)
An equilibrium mixture contains 2.4 mol SO2, 1.2 mol O2 and 0.4 mol SO3. The total pressure is 250 atm. What is the partial pressure of SO3?
25 atm
A buffer solution is prepared by mixing 200 cm3 of 2.00 mol dm–3 propanoic acid, CH3CH2COOH, with 600 cm3 of 1.00 mol dm–3 sodium propanoate, CH3CH2COONa.
Ka for CH3CH2COOH = 1.32 × 10–5 mol dm–3
What is the pH of the buffer solution?
5.06
The table below shows standard entropies, Sө.
CO(g) H2(g) CH3OH(l)
197.6 130.6 239.7
What is the entropy change, ∆Sө, in J mol–1 K–1, for the following reaction?
CO(g) + 2H2(g) CH3OH(l)
-219.1
The redox equilibria for a hydrogen–oxygen fuel cell in alkaline solution are shown below.
2H2O(l) + 2e– <–> H2(g) + 2OH–(aq)
Eө = –0.83 V
1/2O2(g) + H2O(l) + 2e– <–> 2OH–(aq)
Eө = +0.40 V
What is the equation for the overall cell reaction?
H2(g) + 1/2O2(g) –> H2O(l)
Which enthalpy change(s) is/are endothermic?
1. The bond enthalpy of the C–H bond
2. The second electron affinity of oxygen
3. The standard enthalpy change of formation of magnesium
Only 1 and 2
Which statement(s) explain(s) why reaction rates increase as temperature increases?
1. The activation energy is less.
2. Collisions between molecules are more frequent.
3. A greater proportion of molecules have energy greater than the activation energy
Only 2 and 3
Which statement(s) is/are correct for the complex Pt(NH3)2Cl2?
1. One of its stereoisomers is used as an anti-cancer drug.
2. It has bond angles of 109.5°.
3. It has optical isomers.
Only 1
Explain what is meant by the term enthalpy change of hydration.
Enthalpy change when 1 mole of gaseous ions are dissolved in water from hydrated ions to form 1 mole of hydrated ions.
Predict how the enthalpy changes of hydration of F– and Cl – would differ. Explain your answer.
- ∆hydH (F–) is more negative/exothermic than ∆hydH (Cl–)
- F– is smaller in size than Cl–
- F– have greater attraction to H2O
Explain what is meant by the term average bond enthalpy.
Breaking of one mole of bonds in gaseous molecules
Fluorine reacts with steam as shown in the equation below.
2F2(g) + 2H2O(g) –> O2(g) + 4HF(g)
∆H = −598 kJ mol–1
Average bond enthalpies are shown in the table.
Bond Average bond enthalpy / kJ mol–1
O–H +464
O=O +498
H–F +568
Calculate the bond enthalpy of the F–F bond.
- 1856 = 4 x 464 (kJmol–1)
- 2770 (kJ mol–1) = 498 + (4 x 568 (kJ mol–1))
- F–F = (2770 – 1856 – 598) / 2 = (+)158 (kJ mol–1)
This question is about reaction rates. Aqueous iron(III) ions, Fe3+(aq), react with aqueous iodide ions, I–(aq), as shown below.
2Fe3+(aq) + 2I–(aq) –> 2Fe2+(aq) + I2(aq)
A student carries out three experiments to investigate how different concentrations of Fe3+(aq) and I–(aq) affect the initial rate of this reaction. The results are shown below.
Experiment: [Fe3+(aq)]/mol dm–3 —- [I–(aq)] / mol dm–3 —- Initial rate/ mol dm–3 s–1
1: 4.00 × 10–2 —- 3.00 × 10–2 —- 8.10 × 10–4
2: 8.00 × 10–2 —- 3.00 × 10–2 —- 1.62 × 10–3
3: 4.00 × 10–2 —- 6.00 × 10–2 —- 3.24 × 10–3
Determine the rate constant and a possible two-step mechanism for this reaction that are consistent with these results.
- Fe3+ 1st order AND I –2nd order
- k = 8.10 x 10^-4 / (4.00 x 10^2) x (3.00 x 10^-2)^2 = 22.5 dm6 mol–2s-1
- Fe3+(aq) + 2I–(aq) –> Fe+ + I2 SLOW
- Fe3+(aq) + Fe+ 2Fe2+(aq) FAST
Nitrogen monoxide, NO, and oxygen, O2, react to form nitrogen dioxide, NO2, in the reversible
reaction shown in equilibrium 18.1.
Equilibrium 18.1: 2NO(g) + O2(g) <–> 2NO2(g)
Write an expression for Kc for this equilibrium and state the units.
- Kc = [NO2]2 / [NO]2 x [O2]
- Units = dm3 mol–1
A chemist mixes together nitrogen and oxygen and pressurises the gases so that their total gas volume is 4.0 dm3.
- The mixture is allowed to reach equilibrium at constant temperature and volume.
- The equilibrium mixture contains 0.40 mol NO and 0.80 mol O2.
- Under these conditions, the numerical value of Kc is 45.
Calculate the amount, in mol, of NO2 in the equilibrium mixture.
- [NO] = 0.40/4.0 = 0.1(0) (mol dm–3)
- [O2] = 0.80/4.0 = 0.2(0) (mol dm–3)
- [NO2]2 = 45 x 0.10^2 x 0.20 = 0.09
- [NO2] = √(45 x 0.10^2 x 0.20) = 0.3 (mol dm–3)
- Amount NO2 = 0.30 x 4 = 1.2 (mol)
The values of Kp for equilibrium 18.1 at 298 K and 1000 K are shown below.
Equilibrium 18.1: 2NO(g) + O2(g) 2NO2(g)
Temperature / K Kp / atm–1
298 Kp = 2.19 × 1012
1000 Kp = 2.03 × 10–1
i) Predict, with a reason, whether the forward reaction is exothermic or endothermic.
- Exothermic
- Kp decreases as temperature increases
The values of Kp for equilibrium 18.1 at 298 K and 1000 K are shown below.
Equilibrium 18.1: 2NO(g) + O2(g) 2NO2(g)
Temperature / K Kp / atm–1
298 Kp = 2.19 × 1012
1000 Kp = 2.03 × 10–1
ii) The chemist increases the pressure of the equilibrium mixture at the same temperature. State, and explain in terms of Kp, how you would expect the equilibrium position to change.
- Equilibrium position shifts to the right towards the products
- Ratio in Kp expression decreases
- Ratio in Kp expression increases to restore Kp
This question is about acids and bases found in the home. Ethanoic acid, CH3COOH, is the acid present in vinegar. A student carries out an experiment to determine the pKa value of CH3COOH.
- The concentration of CH3COOH in the vinegar is 0.870 mol dm–3.
- The pH of the vinegar is 2.41.
i) Write the expression for the acid dissociation constant, Ka, of CH3COOH.
Ka = [H+] [CH3COO–] / [CH3COOH]
