Question

Menthanol, CH3OH, formally know as wood alcohol is manufacutred commercially by the following reaction: CO (g) + 2H2 (g) --> CH3OH (g) a 1.5000 -L vessel was filled with 0.1500 mol CO and 0.3000 mol H2. When this mixture came to equlilbrium at 500K, the vessel contained 0.1187 mol CO. How many coles of each substance were in the vessel?

Part 2. Obtain the vaule of Kc for the following reaction at 500K

CO (g) + 2H2 (g) --> CH3OH (g) Use the data calculated above.

please show work. thanks

Answer 1

PArt A :

The reaction is

       CO (g)       +          2H2 (g)     -->     CH3OH (g)

Initial                      0.15 / 1.5              0.3 / 1.5             0

Change              -x                             -2x                      x

Equilibrium           0.1-x                       0.2-2x                x

Given x = 0.1187

So moles of CO = 0.15-0.1187 = 0.0313

Moles of H2 = 0.3 - 2X0.1187 = 0.0626

Moles of CH3OH = 0.1187

Part 2

Kc = [CH3OH] / [CO][H2]²

Kc = 0.1187 / 0.0313 X (0.0626)²

Kc = 967.73