In the first part of this exercise we determine the calorimeter constant, Ccal.
ID: 548551 • Letter: I
Question
In the first part of this exercise we determine the calorimeter constant, Ccal. We measure the temperatures of samples of cold and hot water for several minutes. At 3.0 minutes, we mix the water samples and follow the temperature of the resultant mixture for 7 minutes. The hot and cold water temperatures are extrapolated forward to the time of mixing (3 minutes). The temperature of the mixture at the time of mixing (3 min) is extrapolated backward from the data. We do this graphically. The cold water is initially in the calorimeter and therefore the calorimeter undergoes the same temperature change as the cold water. A graph showing the idealized behavior of the system is shown below. The temperature axis is intentionally unlabeled and the data relevant to the determination of the calorimeter constant are in the table below the graph.Explanation / Answer
Part A Determination of calorimeter constant
Density of water =1 g/ml, m = v x d, so 54.1 ml = 54.1 g
Energy lost by the hot water:
q = m Cp T
q = (54.1 g) (4.184 J/g-1 °C-1) (14.6 °C)
q = 3304.77 J
Energy gained by the cold water:
q = m Cp T
q = (47.7 g) (4.184 J/g-1 °C-1) (13.7 °C)
q = 2734.20 J
The calorimeter got the rest:
3304.77 J - 2734.20 J = 570.57 J
Heat capacity of the calorimeter =570.57 J/13.7 °C = 41.65J/°C
Ccal = 41.65J/°C
The equation for the reaction is
HCl + NaOH ------> NaCl + HO
Moles of HCl = 0.050 L HCl × 2.42 mol HCl/1L HCl = 0.121 mol HCl
Moles of NaOH = 0.055 L NaOH × 2.21mol NaOH /1L NaOH = 0.122 mol NaOH
Moles of limiting reagent reacted = 0.121 mol HCl
Volume of solution = (50.0 + 55.0) mL = 105 mL
Mass of solution = 105.0 ml × 1.02g/ml = 107.1 g
T=T2–T1 = (42.1 – 27.6) °C = 14.5 °C
The heats involved are
Heat from neutralization + heat to warm solution + heat to warm calorimeter = 0
nH + mcT + CT = 0
(0.121 mol H) + (107.1 g × 3.97 J·g¹°C¹ × 14.5 °C) + (41.65 J°C¹ × 14.5 °C)= 0
0.121 mol × H + 6165.21 J + 603.93 J = 0
0.121 mol × H = -6769.14 J
H = -6769.14 J/0.121 = -55943 J = - 55.9kJ
Percent of total enthalpy absorbed by calorimeter = (603.93J/55943J) x 100 = 1.08%
Net ionic equation
H+ (aq) + OH- (aq) ----> H2O (l) H = - 55.9kJ
Weak monoprotic acid propionic acid
NaOH ionizes completely as:
NaOH (aq) ----> Na+ (aq) + OH- (aq)
C2H5COOH (aq) <-----> C2H5COO (-) (aq) + H+(aq)
When H+ of acid combine with OH- of base the equilibrium shifts to right and more of propionic acid dissociates. Some of heat produced during combination of H+ and OH- ions is used up for complete dissociation of propionic acid. The heat, which is used up is called enthalpy of dissociation or enthalpy of ionization.
Enthalpy of dissociation of propionic acid = (55.9kJ/mol) - (48.3kJ/mol) = 7.6 kJ/mol
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