Question

Be sure to answer all parts. The equilibrium constant Kc for the reaction H2(g) + CO2(g) ⇌ H2O(g) + CO(g) is 4.2 at 1650°C. Initially 0.72 mol H2 and 0.72 mol CO2 are injected into a 4.5−L flask. Calculate the concentration of each species at equilibrium. Equilibrium concentration of H2: M Equilibrium concentration of CO2: M Equilibrium concentration of H2O: M Equilibrium concentration of CO: M

Answer 1

For the general reaction:

aA + bB <-> cC + dD ,

where a,b,c and d are the stoichiometric coefficients, A and B are reactants and C and D are products;

Kc =

[c][D]d [A]a[B]6

Only compounds in gaseous (g) or aqueous (aq) state are included in the expression.

[…] denotes concentration in molarity

For the given equation :

H2 (g) + CO2(g) ↔H2O (g) + CO(g)

Kc = = 4.2

[H, 0][co] [HQ][002]

The volume of flask = 4.5 L

Molarity = No. of moles/volume (in L)

So, molarity of H2 = 0.72 moles/ 4.5 L = 0.16 M

Molarity of CO2 = 0.72 moles/ 4.5 L = 0.16 M

Let x M H2 and CO2 react to reach equilibrium.

We can construct the following ICE (Initial, Change, Equilibrium) table :

Concentration (in Molarity)	H2	CO2	H2O	CO
Intial	0.16	0.16	0	0
change	-x	-x	+x	+x
equilibrium	0.16 - x	0.16 -x	x	x

Thus, to get molarity of the species present at equilibrium, we need to find x.

Putting the concentration of the three species at equilibrium into equation for Kc , and solving for x:

Kc = = 4.2

[H, 0][co] [HQ][002]

Or, = 4.2

G)(x) (0.16-x)(0.16-x)

or,

(0.16-1)2 = 4.2

Taking square root on both sides:

= 2.05 0.16 - x

Or, x = 0.328 – 2.05x

Or, x = 0.107

Thus, concentration of H2 at equilibrium = 0.16 – x = 0.16 – 0.107 = 0.053 M

concentration of CO2 at equilibrium = 0.16 – x = 0.16 – 0.107 = 0.053 M

concentration of H2O at equilibrium = x = 0.107 M

concentration of CO at equilibrium = x = 0.107 M