Ageng Bashir 01

ASAS TEKNIK KIMIA
AGENG BASHIR DWI PRAKOSO
1 3 31410107
SOAL 25.13
Physicians measure the metabolic rate of conversion of
foodstuffs in the body by usingtables that list theliters of O 2
consumed per gram of foodstuff. For a simple case, suppose
that glucose reacts.
C6H12O6(Glucose) + 6O2(g)
6H2O(l) + 6CO2(g)
How many liters of O2 would be measured for the reaction of
one gram of glucose (alone) if the conversion were 90%
complete in your body? How many KJ/g of glucose would be
produced in the body? Data: 𝛥𝐻𝑓0 of glucose is – 1260 KJ/g
mol of glucose. Ignore the fact that your body is at 37˚C, and
assume it is at 25˚C.
BASIS :
 Suhu terjadinya reaksi = 25˚C
 Tekanannya 1 atm
 1 gmol glukosa (C6H12o6) bereaksi
Reaksi :
C6H12O6(Glucose) + 6O2(g)
6H2O(l) + 6CO2(g)
A.
V O2/gram glukosa = ..........?
Diket :
T1 = 37˚C = 310 K
T0 = 0 ˚C = 273 k
V0 = 22,4 L
Penyelesaian :
PV = nRT
𝑉1
𝑻𝟏
=
𝑉0
𝑇0
𝑉0 𝑇1 22,4 𝐿 𝑋 310 𝐾
𝑉1 =
=
= 25,4359 𝐿 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝑇0
273 𝐾
𝑉 𝑔𝑙𝑢𝑘𝑜𝑠𝑎 =
25,4359 𝐿 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
1 𝑔𝑚𝑜𝑙 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝐿 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝑥
= 0,1413
1 𝑔𝑚𝑜𝑙 glukosa
180 𝑔𝑟𝑎𝑚 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝑔𝑟𝑎𝑚 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
6 𝐿 𝑂2
𝐿 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝐿 𝑂2
𝑉 𝑂2 =
𝑥 0,1413
= 0,8478
1 𝐿 glukosa
𝑔𝑟𝑎𝑚 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝑔𝑟𝑎𝑚 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
B. ΔH = ..........? (kj/gram glukosa)
ΔH = ΔH produk – ΔH reaktan
Penyelesaian :
N C6H12O6 = 1 gmol
N O2 =
6 𝑔𝑚𝑜𝑙 O2
1 𝑔𝑚𝑜𝑙 C6H12O6
𝑥 1 𝑔𝑚𝑜𝑙 𝑐6 𝐻12 O6 = 6 gmol
N H 2O =
6 gmol 𝐻2 O
1 𝑔𝑚𝑜𝑙 C6H12O6
𝑥 1 𝑔𝑚𝑜𝑙 𝐶6 𝐻12 𝑂 6 = 6 𝑔𝑚𝑜𝑙
N CO2 =
6 𝑔𝑚𝑜𝑙 𝐶𝑂2
1 𝑔𝑚𝑜𝑙 C6H12O6
𝑥 1 𝑔𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6 = 6 𝑔𝑚𝑜𝑙
ΔH produk = ΔH H2O + ΔH CO2
= ( n H2O x ΔHf H2O ) + ( n CO2 x ΔHf CO2 )
= ( 6 gmol x ( -285,840 kj/gmol )) + ( 6 gmol x ( 393,51 Kj/gmol ))
= - 1715,04 Kj – 2361,06 Kj
= - 4076,1 Kj
ΔH reaktan = ΔH C6H12O6 + ΔH O2
= ( n C6H12O6 x ΔHf C6H12O6 ) + ( n O2 x ΔHf O2 )
= ( 1 gmol x ( -1260 Kj/gmol )) + ( 6 gmol x 0
kj/gmol )
= - 1260 KJ
ΔH = ΔH produk – ΔH reaktan
= -1260 kj – ( - 4076,1 Kj ) = - 2816 Kj
−2816 𝐾𝑗
1 𝑔𝑚𝑜𝑙 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝛥𝐻 =
𝑥
1 𝑔𝑚𝑜𝑙 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
180 𝑔𝑟𝑎𝑚 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
𝐾𝐽
= 15,6444
𝑔𝑟𝑎𝑚 𝑔𝑙𝑢𝑘𝑜𝑠𝑎
Konversi 90%
glukosa)
ΔH = 0,9 x (-15,6444 Kj/gram
= -14,07996 Kj/gram Glukosa
TRIMAKASIH :D