Chemistry 12
posted by Patrick .
given the reacting system: H2 (G) + I2(G) > 2HI(g) Keq= 64
Equal moles of H2, I2, and HI are placed in a 1.0 L container. Use calculations to determine the direction the reaction will proceed in order to reach equilibirum

..........H2 + I2 ==> 2HI
Keq = 64 = (HI)^2/(H2)((I2)\
Try any number, say 1
Qrxn = (1)^2/1*1 = 1
Try 2.
(2^2)/2*2 = 1
Try 3.
3^2/3*3 = 1
Get the picture?
So Q = 1; K = 64. That means (from Q) that the numerator is too small and the denominator is too large (that's the only way you can get a smaller number than 64). So H2 and I2 are too large; HI is too small. Which way MUST it go?