Question

rating first right answer.

Consider the following reaction:

CO(g)+2H2(g)⇌CH3OH(g)

A reaction mixture in a 5.19-L flask at a certain temperature initially contains 26.5 g CO and 2.36 g H2. At equilibrium, the flask contains 8.64 g CH3OH.

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

Answer 1

Step 1: calculate the concentration of CO

Molar mass of CO,

MM = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass(CO)= 26.5 g

use:

number of mol of CO,

n = mass of CO/molar mass of CO

=(26.5 g)/(28.01 g/mol)

= 0.9461 mol

volume , V = 5.19 L

use:

Molarity,

M = number of mol / volume in L

= 0.9461/5.19

= 0.1823 M

Step 2: calculate the concentration of H2

Molar mass of H2 = 2.016 g/mol

mass(H2)= 2.36 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(2.36 g)/(2.016 g/mol)

= 1.171 mol

volume , V = 5.19 L

use:

Molarity,

M = number of mol / volume in L

= 1.171/5.19

= 0.2256 M

Step 3: calculate the concentration of CH3OH

Molar mass of CH3OH,

MM = 1*MM(C) + 4*MM(H) + 1*MM(O)

= 1*12.01 + 4*1.008 + 1*16.0

= 32.042 g/mol

mass(CH3OH)= 8.64 g

use:

number of mol of CH3OH,

n = mass of CH3OH/molar mass of CH3OH

=(8.64 g)/(32.04 g/mol)

= 0.2696 mol

volume , V = 5.19 L

use:

Molarity,

M = number of mol / volume in L

= 0.2696/5.19

= 5.195*10^-2 M

step 4: now calculate Kc