Page 48 - E-LKPD.kesetimbangan kimia

3 mol
[NO ] = = 1,5 M
2
2 l
2 mol
[NO] = = 1 M
2 l
1 mol
[O ] = 2 l = 0,5 M | 47
2

Sehingga,
2
2
[NO] [O ] [1] [0,5] 0,5
2
-1
K = [NO ] 2 = [1,5] 2 = 2,25 = 2,2 X 10
c
2

3. 2 3( ) ⇄ 2 2( ) + 2( )

2
(P SO 2 ) (P O 2 )
Kp = 2
(P SO 3 ) Maka:
2
P total = 10 atm, maka: (P ) (P )
Kp = SO 2 O 2
mol SO 2 (P ) 2
P SO = × total SO 3
 2 mol total (6) (2)
2
0,3 mol Kp = atm
P SO = 0,5 mol × 10 atm (2) 2
2
P SO = 6 atm Kp = 72 atm
2
4
Kp = 18 atm

mol 2
 P O = × total
2
mol total
0,1 mol
PO = × 10 atm
2
0,5 mol
P O = 2 atm
2

mol 3
 P SO = mol total × total
3
0,1 mol
P SO = × 10 atm
3
0,5 mol
P SO = 2 atm
3
C. PERTEMUAN 3

Hipotesis

Konsentrasi mempengaruhi tekanan parsial gas berdasarkan hubungan
tekanan (P) dengan konsentrasi menggunakan persamaan gas ideal,

yaitu : PV = n R T