AS Paper 1 A-Level Chemistry Flashcards
This question is about ionisation energies of Group 2 elements. Explain why the first ionisation energy of the Group 2 elements decreases down the group.
the outer electron is in a higher (energy) level / there is an increase in shielding / the atoms get larger / more shells There is a weaker attraction between the nucleus and the outer electron.
Give an equation, including state symbols, to represent the process that occurs when the third ionisation energy of magnesium is measured.
Mg2+(g) → Mg3+(g) + e–
Explain why the third ionisation energy of magnesium is much higher than the second ionisation energy of magnesium.
The electron is removed from 2p subshell / 2nd energy level / lower energy level / subshell that is closer to the nucleus (Electron being removed is) less shielded (than 3s)
This question is about acid–base titrations. Citric acid reacts with sodium hydroxide.
C6H8O7(aq) + 3 NaOH(aq) → Na3C6H5O7(aq) + 3 H2O(l)
A student makes a solution of citric acid by dissolving some solid citric acid in water. Describe a method to add an accurately known mass of solid to a beaker to make a solution.
M1 measure the mass of the weighing boat (or similar) and solid M2 Add the solid to a beaker (or other suitable container) and then reweigh the weighing boat (and subtract to find the mass of solid added.) OR M1 Place weighing boat on a balance and zero the balance M2 Add the solid to a beaker (or other suitable container), wash out weighing boat and transfer washing to the beaker.
The student dissolves 0.834 g of citric acid in water and makes the solution up to 500 cm3 Calculate the concentration, in mol dm–3, of citric acid in this solution.
M1 Mr citric acid = 192.0 M2 Amount of citric acid = Mass / Mr = 0.834 / 192 = 0.0043438 (mol) M3 Concentration = moles / volume = 0.0043438 / 0.5 = 0.00869 (mol dm–3)
Hydrogen can be prepared on an industrial scale using the reversible reaction
between methane and steam.
CH4(g) + H2O(g) ⇌ CO(g) + 3 H2(g) ΔH = +206 kJ mol−1
The reaction is done at a temperature of 800 °C and a low pressure of 300 kPa
in the presence of a nickel catalyst.
Explain, in terms of equilibrium yield and cost, why these conditions are used. [6]
Stage 1: Temperature
1a. The reaction is endothermic (so equilibrium shifts to right hand side to reduce the temperature)
1b. So, higher temperature increases the yield
1c. High temperatures are costly (so compromise temperature used)
Stage 2: Pressure
2a. More moles of gas on the right hand side, (so equilibrium shifts to right hand side to increase the yield)
2b. So, lower pressure increases the yield
2c. A low pressure means a low cost
Stage 3: Catalyst
3a. Catalyst has no effect on yield
3b. Adding a catalyst allows a lower temperature to be used
3c. So, this lowers the cost
Which statement about the use of a catalyst in a reversible reaction is correct?
A The activation energy for the reverse reaction is
increased.
B The equilibrium constant increases.
C The rate of the reverse reaction increases.
D The enthalpy change for the forward reaction
decreases. [1]
C
This question is about the equilibrium
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
(a) State and explain the effect, if any, of a decrease in overall pressure on the
equilibrium yield of SO3
Effect:
Explanation:
[3]
M1 decreases yield
M2 So equilibrium shifts to side with more moles/molecules or
more moles/molecules on left hand side
Allow M2 independent of M1
M3 So equilibrium shifts (to left side) to oppose decrease in
pressure OR to increase pressure
Must refer to equilibrium shifting to gain maximum
marks
This question is about equilibria.
Give two features of a reaction in dynamic equilibrium. [2]
forward and reverse reactions proceed at equal rates concentrations (of reactants and products) remain constant or concentrations (of reactants and products) stay the same
allow answers in either order do not accept equal concentrations do not accept concentrations are the same ignore closed system
A gas-phase reaction is at equilibrium. When the pressure is increased the yield of product decreases. State what can be deduced about the chemical equation for this equilibrium. [1]
more moles of (gaseous) products (than (gaseous) reactants) or more moles on the RHS (than LHS)
allow molecules do not accept atoms
Rhenium has an atomic number of 75
Define the term relative atomic mass. [2]
average/mean mass of 1 atom (of an element) 1/12 mass of one atom of 12C
or
average/mean mass of atoms of an element 1/12 mass of one atom of 12C
or
average/mean mass of atoms of an element ×12 mass of one atom of 12C
or
(average) mass of one mole of atoms 1/12 mass of one mole of 12C or
(weighted) average mass of all the isotopes 1/12 mass of one atom of 12C
or
average mass of an atom/isotope (compared to C−12) on a scale in which an atom of C−12 has a mass of 12
Additional comments/guidelines
M1 = top line M2 = bottom line if moles and atoms/isotopes mixed max = 1
The relative atomic mass of a sample of rhenium is 186.3 Table 2 shows information about the two isotopes of rhenium in this sample.
Table 2:
Relative isotopic mass Relative abundance
185 10
To be calculated 17
Calculate the relative isotopic mass of the other rhenium isotope. Show your working. [2]
State why the isotopes of rhenium have the same chemical properties. [1]
M1 186.3 = (185 × 10) + (X × 17) / 27
M2 (relative isotopic mass) = 187(.1)
same electron configuration
allow same number of electrons allow same electron structure ignore same number of protons ignore different number of neutrons do not accept same number of neutrons
A sample of rhenium is ionised by electron impact in a time of flight (TOF) mass spectrometer.
A 185Re+ ion with a kinetic energy of 1.153 × 10−13 J travels through a 1.450 m flight tube.
The kinetic energy of the ion is given by the equation KE = 1/2 mv^2 where
m = mass / kg
v =speed / m s–1
KE = kinetic energy / J
Calculate the time, in seconds, for the ion to reach the detector. The Avogadro constant, L = 6.022 × 1023 mol−1 [5]
M1 mass 185Re = 184/6.02 × 10^23 × 1000 = 3.072 x 10^-25 (kg)
M2 v = d/t
M3 v^2 = 2KE / m or 7.5(0) × 10^11
M4 v = √2KE/m or 8.66 × 10^5
M5 t = 1.45 / 8.66 × 10^5 = 1.67 × 10^–6 (s)
Additional comments/guidelines
calculate mass in kg
recall of v = d/t
rearrangement to get v^2
allow √2 × 1.153 × 10–13 / M1
M5 t = 1.45 / M4
allow 1.67 × 10^-6 to 1.68 × 10^–6 (s)
Oxygen has a boiling point of -183°C but hydrogen chloride has a much similar Mr but a higher boiling point of -85°C. Explain why. [3]
HCl has permanent dipole-dipole forces whereas oxygen has van der waal’s forces. The dipole-dipole forces are stronger and therefore require more energy to overcome.
Ethanol, CH3CH2OH boils at 78°C while a compound with the same molecular formula methoxymethane, CH3OCH3 boils at -25°C. Explain why. [3]
Ethanol has hydrogen bonds whereas methoxymethane has permanent dipole-dipole forces. The hydrogen bonds are stronger and therefore required more energy to overcome.
