Question

Consider the following reaction: CO (g) + H2O (g) ⇌ CO2 (g) + H2(g) If you start with a mixture containing 1.00 mol of CO and 1.00 mol of H2O, calculate the number of moles of each component in the mixture when equilibrium is reached at 1000 °C. The mixture contains 0.43 mol H2?

nCO =

nH2O =

nCO2 =

How do I work this out?

Answer 1

Given:-

Initial no. of moles of CO = 1.00 mol

Initial no. of moles of H2O = 1.00 mol

Equilibrium no. of moles of H2, nH2 = 0.43 mol

Given chemical reaction:-

CO (g) + H2O (g) ⇌ CO2 (g) + H2(g)

Filling the ICE Table for the above reaction for given data-

	CO(mol)	H2O(mol)	CO2(mol)	H2(mol)
I	1.00	1.00	0	0
C	-x	-x	+x	+x
E	1.00 - x	1.00 - x	x	x

Here,

E = I + C

I stands for the initial no. of moles of components.

C stands for change in no. of mole of components.

E stands for equilibrium no. of moles of components.

-ve sign denotes the consumption.

+ve sign denotes the production.

From the ICE Table,

Equilibrium no. of moles of H2, nH2 = x

= 0.43 mol ------given

Therefore, x = 0.43 mol

Equilibrium no. of moles of CO, nCO = (1.00 - x) mol

= (1.00 - 0.43) mol

= 0.57 mol

Equilibrium no. of moles of H2O, nH2O = (1.00 - x) mol

= (1.00 - 0.43) mol

= 0.57 mol

Equilibrium no. of moles of CO2, nCO2 = x mol

= 0.43 mol

Hence, no. of moles of each component when equilibrium is reached are as follows-

nCO = 0.57 mol

nH2O =  0.57 mol

nCO2 = 0.43 mol