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e. Penentuan kalor reaksi Zn + CuSO4
Zn(s) + CuSO4 ZnSO4(aq) + Cu(s)
Massa larutan = ρ x V = 1.14 x 40 = 45.60 gram
Ar Zn= 65
T CuSO4 = 27.5° C = 300.5 K
M CuSO4 = 0.5 M
V CuSO4 = 40 mL = 0.04 L
n CuSO4 = M x V
= 0.5 x 0.04
= 0.02 mol
Kalor yang diserap calorimeter (Q1) = C x ΔT
= 151.2 (303-298)
= 756 J
Kalor yang diserap larutan (Q2) = m x c x ΔT
= 45.6 x 3.52 x (303-298)
= 802.56 J
Kalor yang dihasilkan oleh reaksi(Q3) = Q1+Q2
= 756 + 802.56
= 1558.56
Kalor reaksi (Q4) = Q 3mol
= 1558.560.02
= 77928 J/mol
f. Penentuan kalor pelarutan etanol dalam air
● Komposisi A
V air = 2 mL
V etanol = 8 mL
Mr etanol = 46
n etanol = metanolMr
= 6.34446
= 0.138 mol
ρ air = 1 gr/mL
ρ etanol = 0.793 gr/mL
c air = 4.2 J/gK
c etanol = 1.92 J/gK
T air = 26° C = 299 K
T etanol = 25° C = 298 K
T konstan = 28°C = 301 K
ΔTa = T konstan – T air
= 301 – 299
= 2 K
ΔTc = T konstan – Tc
= 301 – 298
= 3 K
m air = ρ air x V air
= 1 gr/mL x 2 mL
= 2 gram
m etanol = ρ etanol x V etanol
= 0.793 gr/mL x 8 mL
= 6.344 gram
Q7 = Qair
= ma x ca x ΔTa
= 2 x 4.2 x 2
= 16.8 J
Q8 = Qetanol
= me x ce x ΔTe
= 6.344 x 1.92 x 3
= 36.54 J
Q9 = K x ΔTa
= 214.188 x 2
= 428.376 J
Q10 = Q7 + Q8 +Q9
=16.8 + 26.54 + 428.376
= 481.716 J
ΔH = Q 10netanol
= 481.7160.138
= 3490.699 J/mol
= 3.491 KJ/mol
Komposisi B
Vair= 4 mlV etanol= 6 mlM air = ρ air x V air
= 1 gr/ml x 4 ml=4 gr
M etanol = ρ etanol x V etanol= 0.793 x 6=4.758 gr
Tkonstan = 27.75°C = 300.75 KΔTa = Tkonstan – Tair = 300.75 – 299
= 1.75 KΔTe = Tkonstan – Tetanol
= 300.75 – 298 = 2.75 K
Mol etanol = mMr
= 4.75846
= 0.103 mol
Q7 = Qair
= ma x ca x ΔTa
= 4 x 4.2 x 1.75
= 29.4 J
Q8 = Qetanol
= me x ce x ΔTe
= 4.758 x 1.92 x 2.75
= 25.121 J
Q9 = K x ΔTa
= 214.188 x 1.75
= 374.829 J
Q10 = Q7 + Q8 +Q9
=29.4 + 25.121 +374.829
= 429.35 J
ΔH = Q 10netanol
= 429.350.103
= 4168.447 J/mol
= 4.168 KJ/mol
Komposisi C
Vair= 6 mlV etanol= 4 mlM air = ρ air x V air
= 1 gr/ml x 6 ml=6 gr
M etanol = ρ etanol x V etanol= 0.793 x 4=3.172 gr
Tkonstan = 28+27.75
2 = 27.875°C = 300.875 K
ΔTa = Tkonstan – Tair = 300.875 – 299
= 1.875 KΔTe = Tkonstan – Tetanol
= 300.875 – 298 = 2.875 K
Mol etanol = mMr
= 3.17246
= 0.069 molQ7 = Qair
= ma x ca x ΔTa = 6 x 4.2 x 1.875 = 47.25 J
Q8 = Qetanol= me x ce x ΔTe= 3.172 x 1.92 x 2.875= 17.509 J
Q9 = K x ΔTa
= 214.188 x 1.875
= 401.602 J
Q10 = Q7 + Q8 +Q9=47.25 + 17.509 + 401.602= 466.361 J
ΔH = Q 10netanol
= 466.360.069
= 6758.855 J/mol= 6.759 KJ/mol
Komposisi D
Vair= 8 mlV etanol= 2 mlM air = ρ air x V air
= 1 gr/ml x 8 ml=8 gr
M etanol = ρ etanol x V etanol= 0.793 x 2=1.568 gr
Tkonstan = 28+27.75
2 = 27°C = 300 K
ΔTa = Tkonstan – Tair = 300 – 299
= 1 KΔTe = Tkonstan – Tetanol
= 300 – 298 = 2 K
Mol etanol = mMr
= 1.58646
= 0.034 molQ7 = Qair
= ma x ca x ΔTa = 8 x 4.2 x 1 = 33.6 J
Q8 = Qetanol= me x ce x ΔTe= 2 x 1.92 x 2= 7.68 J
Q9 = K x ΔTa= 214.188 x 1= 214.188 J
Q10 = Q7 + Q8 +Q9=33.6 + 7.68 + 214.188
= 255.468 J
ΔH = Q 10netanol
= 255.4680.034
= 7513.7647 J/mol= 7.513 KJ/mol
