# Question 16 (Mandatory) (4.256 points) Carbonyl bromide decomposes to carbon monoxide and bromine. COBr2(g) = CO(g)... Question 16 (Mandatory) (4.256 points) Carbonyl bromide decomposes to carbon monoxide and bromine. COBr2(g) = CO(g) + Br2(g) Kc is 0.19 at 73 °C. If an initial concentration of 0.63 M COBr2 is allowed to equilibrate, what are the equilibrium concentrations of COBr2, CO, and Br2? Oa) (COBr2] = 0.30 M, [CO] = 0.33 M, (Bra] = 0.33 M Ob) (COBr2] = 0.63 M, [CO] = 0.35 M, (Br2] = 0.35 M Oc) [COBr2] = 0.11 M, [CO] = 0.26 M, (Br2] = 0.26 M d) [COBr2] = 0.28 M, [CO] = 0.35 M, (Br2] = 0.35 M e) (COBr2] = 0.37 M, [CO] = 0.26 M, (Br2] = 0.26 M Question 17 (Mandatory) (4.256 points)

Correct option is : (e) [COBr2] = 0.37 M, [CO] = 0.26 M, [Br2] = 0.26 M

Explanation

Given : initial concentration of COBr2 = 0.63 M

Kc = 0.19

 ICE table COBr2 (g) CO (g) Br2 (g) Initial conc. 0.63 M 0 0 Change -x +x +x Equilibrium conc. 0.63 M - x +x +x

Kc = [CO]eq[Br2]eq / [COBr2]eq

0.19 = [(x) * (x)] / (0.63 M - x)

Solving for x, x = 0.264 M

[CO]eq = [Br2]eq = x = 0.26 M

[COBr2]eq = 0.63 M - x

[COBr2]eq = 0.63 M - 0.26 M

[COBr2]eq = 0.37 M

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