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Soal Termokimia No 21-30

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21. 5 L air dipanaskan hingga suhu air tersebut naik dari 27 C menjadi 40 C. Jika diketahui massa jenis air = 1gr/ml dan kalor jenis air = 4,2 J. Maka tentukanlah  ⧍H reaksi pemanasan tersebut !

→ Dik : V = 5000 ml
              T1 = 27 C
              T2 = 40 C
              C = 4,2

Maka Q = m.c.⧍T
              = 5000 . 4,2 . ( 40 -27 )
              = 273.000 = 273 Kj


22. Diketahui :

⧍Hf CO2 = -393,5 Kj
⧍Hf H2O = -242 Kj

CH4 + 2O2 → 2H2O + CO2  ⧍H = -677 Kj

Maka besarnya ⧍Hf CH4 adalah.....

→ Kita cari dengan cara berikut !


23. Jika diketahui ⧍H pembentukan gas karbon dioksida ( CO2 ) , uap air ( H2O ) dan gas propana berturut-turut yaitu -393,5 Kj, -242 Kj, dan 104 Kj. Tentukan banyak panas yang dibebaskan pada pembakaran 100 gr propana !

→ Pertama, Kita buat proses reaksinya terlebih dahulu !

C3H8 + 5O2 → 3CO2 + 4H2O

Kedua, kita cari ⧍H dari pembakaran 1 mol propana tersebut !


24. Dalam suatu reaksi kimia dibebaskan 100 Kj energi. Jika kalor ini digunakan untuk memanaskan 100 ml air, maka kenaikan suhunya ialah ? ( kalor jenis air = 4,2 )

⧍T = 100.000 / 4,2 x 100 = 100.000 / 420 = 238 C

Jadi kenaikan suhunya ialah sebesar 238 C


25. Ke dalam kalorimeter dituangkan 100 gr air panas ( 100 C ), kemudian ditambahkan 75 gr air dingin ( 10 C ) sehingga suhu campuran menjadi 40 C. Jika suhu kalorimeter naik sebesar 10 C , tentukan kapasitas kalor kalorimeter ? Diketahui kalor jenis air = 4,2

→ Jumlah kalor dilepaskan air panas sama dengan jumlah kalor yang diserap oleh air dingin dan kalorimeter

Qair panas = Qair dingin + Qkalorimeter

Qkalorimeter = c. ⧍T

Qair panas = 100 x 4,2 x ( 40 -100 )
                  = -25200 J

Qair dingin = 75 x 4,2 x ( 30 )
                   = +9450

= Qair panas + Qair dingin + Qkalorimeter
= -25200 + 9450 + Ckalorimeter x 10 = 0
Ckalorimeter = 34650 / 10 = 346,5 J

Jadi kapasitas kalor dari kalorimeter ialah sebesar 346,5 J



26. Dalam sebuah kalorimeter, direaksikan 50 gr HCl dengan 5 gr logam magnesium. Jika suhu asal campuran 27 C dan setelah reaksi suhunya 50 C, maka tentukanlah ⧍H reaksinya ! Diketahui kalor jenis = 3 J/g, kapasitas kalor kalorimeter = 150 J/g dan suhu kalorimeter naik sebesar 5 C

→ Jumlah kalor yang terlibat pada reaksi tersebut ialah sebanyak :

Qreaksi + Qlarutan + Qkalorimeter = 0

Qreaksi = -( Qlarutan + Qkalorimeter )

Qlarutan = m. C . ⧍T
               = 55 . 3 . 23 = 3795 J

Qkalorimetri = c . ⧍T
                      = 150 . 5 = 750 J

Maka Q reaksi :

Qreaksi = - ( 3795 + 750 ) = -4545 J

Jadi ⧍H reaksi yang dihasilkan ialah sebesar -4545 J


27. Perhatikan reaksi kimia dibawah ini :

C + O2 → CO2  ⧍H = -393,5 Kj
H2 + 1/2 O2 → H2O  ⧍H = -283,5 Kj
2C + H2 → C2H2  ⧍H = -226,5 Kj

Atas dasar diatas, maka kalor reaksi dari reaksi dibawah ini ialah :

C2H2 + 5/2 O2 → H2O + 2CO2

→ Mari kita gunakan prinsip hukum hess untuk menjawab soal diatas !

Jadi besarnya kalor dari reaksi diatas ialah sebesar -843,8 Kj


28. Perhatikan reaksi berikut !

C + O2 → CO2  ⧍H = -394 Kj
2CO + O2 → 2CO2  ⧍H = -569 Kj

Reaksi pembentukan 40 gr gas CO ( mr = 28 ) disertai dengan ⧍H sebesar ?

→ Kita gunakan lagi prinsip hukum hess :

Maka kalor yang dilepaskan pada pembentukan 140 gr gas CO ialah sebanyak :

140/28 x 109,5 = -547,5 Kj


29. Jika kalor netralisasi ialah 120 kkal/mol , maka kalor netralisasi 100 ml HCl 0,1 M dan 150 ml 0,075 M NaOH ialah...

→ Pertama kita buat reaksi netralisasi nya dan kita cari juga jumlah mol masing-masing senyawa !

HCl + NaOH → NaCl + H2O

mol HCl =  100 x 0,1 = 10 mmol
mol NaOH = 150 x 0,075 = 11,25 mmol

Jumlah mol yang terbatas ialah HCl karena memiliki mol paling kecil yaitu 0,01 maka ⧍H nya ialah sebesar :

⧍H = 0,01 x 120 = 1,2 kkal


30. Perhatikan reaksi dibawah ini :

a. O3 + Cl → O2 + ClO  ⧍H = -120 Kj
b. ClO + O → O2 + Cl  ⧍H = -270 Kj
c. O3 + O → 2O2

nilai ⧍H dari reaksi yang terakhir ialah sebesar ...

→ Reaksi C merupakan penjumlahan dari reaksi A dan B, maka jumlah kalor dari reaksi C ialah sebesar :

>> Selanjutnya ( Soal No 31-40 )

300x600
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1 Response to "Soal Termokimia No 21-30"

  1. penjelasannya kurang jelas, jadi suka ada yang saya gak tau itu dapet angka segitu darimana

    ReplyDelete

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