Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations of [CO]=0.27M and [H2]=0.49M. At equilibrium, the concentration of CH3OH is 0.11 M. Find the equilibrium constant at this temperature.

Ka = 4.76108

Explanation:

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

i belive it is water