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Carbonyl bromide, COBr2, decomposes to CO and Br2 by the following simple, gas-phase, reaction: 
COBr2(g) ↔ Br2(g) + CO(g)

If a 1.1 liter flask is found to contain 0.018 mol of COBr2 at equilibrium, what will be the equilibrium concentrations of CO and Br2 in the same flask? The equilibrium constant for this reaction is K= 0.190.

May 8th, 2015

DATA:

Moles of CoBr2 = (0.018 mol*1000)/1100 = 0.0163M

Keq = 0.190

[Co]=?

[Br2]=?

SOLUTION:

Keq = [Br2][Co]/[CoBr2]

Suppose,

X= [Br2]

As,

[Br2]=[Co]

So,

X=x

Put the values in equation;

0.190= (x*x)/0.0163

X2=0.0031

X= 0.0557M

S0, the molar concentration of [Br2] and [Co] is 0.0557M

May 8th, 2015

please enter your answer in the scientific figure

May 8th, 2015

x= 5.57*10^-2M

May 8th, 2015

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