the equilbrium constant, kc for the following(READ BELOW THE QUESTION CONTINUES)?

Question

reaction at 425 C is 54.8
H2(g)+ I2(g) ¬ 2HI(g)
if 0.5 mol dm-3 of both H2 (g) and I(g) are allowed to come to equilbrium at 425 C in a closed vessel, what is the concerntration of all three species at equilibrium?

Answer 1

Let the concentration of hydrogen iodide at equilibrium be x. The concentration of hydrogen and of iodine vapour will be:
0·5 - x
Then you set up the equation:
x² ÷ (0·5 - x)² = 54·8
Taking the square root of both sides (+ve roots only):
x ÷ (0·5 - x) = 7·40
From this, you'll get:
x = 0·440
and 0·5 - x = 0·0596

Hence, concentration of hydrogen iodide at equilibrium will be 0·440 mol/dm³
Concentration of hydrogen and that of iodine are both
0·0596 mol/dm³.

It should be pointed out that this whole thing is a bit impractical, since experiments on gas-phase equilibria are usually carried out at fairly low pressures, as it's difficult to get reproducible results at high pressures. I reckon the reaction vessel here would have to be at 55 to 60 atm.

Answer 2

All you need to do is to solve the equation x squared/(0.5 - x) squared = 54.8.