Time, min Temperature, ° C Time, min Temperature, ° C Mixture HCl NaOH 0 23.1 3
ID: 901007 • Letter: T
Question
Time, min
Temperature, °C
Time, min
Temperature, °C
Mixture
HCl
NaOH
0
23.1
3
6
35.3
0.5
23.15
2.9
7
35.8
1.0
23.2
2.8
8
36.3
1.5
23.25
2.8
9
36.1
2.0
23.3
2.7
10
36.05
2.5
23.33
2.7
11
35.95
3.0
23.35
2.7
12
35.9
3.5
23.35
2.7
13
35.75
4.0
23.4
2.7
14
35.75
4.5 (T initial)
23.4
2.7
15
35.75 (T final)
5.0
mixing
1. Plot the temperature-time data and determine the mean temperature of the unmixed reagents.
Time, min
Temperature, °C
Time, min
Temperature, °C
Mixture
HCl
NaOH
0
23.1
3
6
35.3
0.5
23.15
2.9
7
35.8
1.0
23.2
2.8
8
36.3
1.5
23.25
2.8
9
36.1
2.0
23.3
2.7
10
36.05
2.5
23.33
2.7
11
35.95
3.0
23.35
2.7
12
35.9
3.5
23.35
2.7
13
35.75
4.0
23.4
2.7
14
35.75
4.5 (T initial)
23.4
2.7
15
35.75 (T final)
5.0
mixing
Explanation / Answer
Mean T of unmixed reagents = (T(reagent1)+T(reagent2))/2=(23.4+2.7)/2=13.1^0 C
T=T_final-T_initial=35.75-13.1=22.65^0 C
q_(reac.mix)=(V_HCl+V_NaOH )*d*C_(reac.mix) T=100*1.04*3.89*22.65=9163.284 J
q_cal=C_cal*T=46.6*22.65=1055.49 J
q_tot=q_(reac.mix)+q_cal=9163.284+1055.49=10218.774 J
q_(released by reaction)=q_tot=10218.774 J
The limiting reagent is HCl because it is the smallest No. of moles and upon it the liberated quantity of heat determined
H_(neut.)=q_tot/n_(limiting reagent) =q_tot/((V_HCl*M_HCl/1000))=10218.774/((50*2.00297/1000))=10218.774/0.1001485=102036.216 J/mol
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