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
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 =
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
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
Or, = 4.2
or,
Taking square root on both sides:
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
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