Q13.
(a) M1 Temperature on the y-axis and uses sensible scales (i.e. minimum 20 little squares for each °C on y-axis)
Lose mark if temperature scale starts at 0°C
This mark scores if all points fit on the grid. Do not penalise M1 if extrapolation to 4 mins goes off the grid – this is penalised in M3.
1
M2 Plots all the points correctly (± half a small square)
Lose mark if the points go off the grid Ignore a plotted point at 4 mins used to work out ∆T
1
Both lines must be straight and through all points except 5th minute; lose mark if the lines are kinked/doubled. Any line through 5th minute loses mark
“S-shaped curve” through points loses M3 and M4
M3 Draws two best fit lines (0-3 mins) and (6-12 mins)
1
M4 Extrapolates both lines to at least the 4th minute
Lose mark if the extrapolation goes off the grid.
1
M5 21.9 – 19.8 = 2.1 (°C)
Allow calculation of ∆T from S-shaped curve as: Value at 4th minute – 19.8 but not if 0 (°C)
∆T value ecf from incorrect lines/extrapolation
∆T must be to at least 1dp
If value of ∆T = 2.1, then award M5
1
(b) 0.2 / 2.1 × 100 = 9.5 %
Conseq on (a)
Ignore no of sfs.
1
(c) Replace the glass beaker with a polystyrene cup / insulate the glass beaker / use a lid
Ignore use more dilute solutions
Ignore suggested materials for insulation
Do not allow copper calorimeter / bomb calorimeter
1
(d) Increase magnitude of temperature change
Ignore references to volume changes
1
by increasing the concentration of the acid/alkali
Mark independently
1
(e) M1 HOOCCOOH + 2KOH → K2(OOCCOO) + 2H2O
M1 – equation (allow ionic KOH / 2C2O4)
H2C2O4 + 2KOH → K2C2O4 + 2H2O
ignore state symbols
allow multiples
Mark independently
M2 q ( = mc∆T) = 100 × 4.2 × 3.2 = 1344 J
M2 – process
(ignore sign here)
(allow calculations involving 4.18 which leads to 1338 J)
1
M3 n HOOCCOOH = 25 × 0.800 / 1000 = 0.020
n KOH = 75 × 0.6 / 1000 = 0.045
M3 – calculations of amounts, in moles, of both the ethanedioic acid and potassium hydroxide (both calculations needed)
1
M4 Moles of water = 0.040 moles
M4 – answer (stated or used in calculation of ∆H)
1
M5 ∆H = –1.344/0.04
= – 33.6 ( kJ mol–1)
M5 – ecf on M2 and M4
Answer must be negative and to at least 2sf
∆H = – M2 (in kJ) / M4
–32.5 – –34 scores 4/4 (M2–M5 + equation)
+32.5 – +34 scores 3/4 (M2, M3, M4 + equation)
–65 – –68 scores 3/4 (+ equation)
+65 – +68 scores 2/4 (+ equation)
–52 – –54 scores 3/4 (+ equation)
+52 – +54 scores 2/4 (+ equation)
2
(f) HOOCCOOH is a weak acid / not fully dissociated
H2SO4 is a strong(er)acid / fully dissociated / dissociates more
1
(more) energy needed to break bonds/complete dissociation / dissociation is endothermic
So less energy is needed for dissociation of sulfuric acid
Ignore references to heat loss
1[16]