H2(g) + I2(g) ⇌ 2HI(g)
Let us consider the formation of HI in which ‘a’ moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’. Let ‘x’ moles of each of H2, and I2, react together to form 2x moles of HI.
|
H2 |
I2 |
HI |
Initial number of moles |
a |
b |
0 |
Number of moles reacted |
x |
x |
0 |
Number of moles at equlibrium |
a - x |
b - x |
2x |
Active mass |
(a - x) / V |
(b - x) / V |
2x / V |
Applying law of mass action.