Predict and calculate the effect of concentration changes on an equilibrium syst

Question

Predict and calculate the effect of concentration changes on an equilibrium system. Some COBr_2 is allowed to dissociate into CO and Br_2 at 346 K. At equilibrium, [COBr_2] = 0.203 M, and [CO] = [Br_2] = 0.196 M. Additional COBr_2 is added so that [COBr_2]_new = 0.309 M and the system is allowed to once again reach equilibrium. COBr_2(g) doubleheadarrow CO(g) + Br_2(g) K = 0.190 at 346 K. In which direction will the reaction proceed to reach equilibrium What are the new concentrations of reactants and products after equilibrium? [COBr_2] = M [CO] = M [Br_2] = M

Explanation / Answer

COBr2 -------------> CO + Br2

a) The reaction will proceed to the right to reach the equlibirum. Accodring to Le Chatlier's principle any change in the parameters like concentration, temperature and pressue at equlibriumwill shift its equlibirum to counter this effect.

so if you add COBr2, the concentration of product changes by converting the reactants to products and hence driving the reaction to the product side to keep the equlibrium constant value same.

b) K = 0.190

Total amount of COBr2 (after the addition) = 0.309M

K = [CO][Br2]/[COBr2]

let x be the equlibrium concentration of Br2 and CO. K is same since the temperature is maintained at 346K. The amount of COBr2 is 0.309-x.

So the equilbirum for the reaction is

K = (x2)/(0.309-x)

0.190(0.309-x) = x2

0.05871-0.109x = x2

x2 + 0.109x-0.05871 = 0 (Quadratic equation)

solving for x gives 0.165

So the concentration of CO and Br2 at equlibrium after the additon of COBr2 is 0.165M.

The concentration of COBr2 at equlibrium us 0.309-0.165 = 0.144M

[COBr2] = 0.144M

[CO] = 0.165M

[Br2] = 0.165M

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.