Question

Consider the following reaction:
CO(g)+2H2(g)⇌CH3OH(g)
A reaction mixture in a 5.25 −L flask at a certain temperature contains 26.8 g CO and 2.35 g H2. At equilibrium, the flask contains 8.64 g CH3OH

Part A

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answer 1

initial,
[CO] = mass of CO /(molar mass of CO * volume)
= 26.8/(28*5.25)
=0.179 M

[H2] = mass of H2 /(molar mass of H2 * volume)
= 2.35/(2*5.25)
=0.224 M

at equilibrium,
[CH3OH] = mass of CH3OH /(molar mass of CH3OH * volume)
= 8.64/(32*5.25)
=0.0514 M

CO(g) + 2H2(g) <------> CH3OH(g)
0.179      0.224            0     (initial)
0.179-x    0.224-2x         x (at equilibrium)

clealry, x = 0.0514 M

Kc = [CH3OH]/[CO][H2]^2
Kc =x/(0.179-x)(0.224-2x)^2
Kc= 0.0514/{(0.179-0.0514)(0.224 - 2*0.0514)^2}
Kc = 0.0514/(0.1276*0.01469)
   = 27.4

Answer: 27.4