3. The decomposition reaction of COCh is, COCL()Co)+Cl(g) Kc-22x 1010 The initia

Question

3. The decomposition reaction of COCh is, COCL()Co)+Cl(g) Kc-22x 1010 The initial concentration of CoClz is 0.3166 M. (a) At equilibrium, the concentrations of the products are concentration of reactant. the (C) equal to (A) greater than (B) less than (D) unknown because more information is needed to determine. (b) Fill the ICE table below. ICO 0 COCE 12 0 Initial concentration (M) Changes (M) Equilibrium concentration (M) -x (c) Calculate the equilibrium concentration of all species. At equilibrium, [COCb] = ' [CO] = 4. Compare the basicity of the following bases:OH, NH2, oCr, OF, and ClOs. (a) From weakest to strongest base- (b) Assign the Ka values (3.0 x 10, 2.0 x 10", 1.0 o-14, 32 x 10-33, and 10) of the conjugate acids to the above bases based on your answer in (a). Also calculate pKa and pkb. Write your answers in the corresponding cells in the below table. Ka pkapkb Bases Conjugate acid OH- NH2 of CIO3

Explanation / Answer

Q3

a)

in equilbirium, the concentration of PRODUCTS will be LESS than,

since Kc data states this does not forward products

b)

initially

[COCl2] = 0.3166

[CO] = 0

[Cl2] = 0

in equilbirium

[COCl2] = - x

[CO] = + x

[Cl2] = x

in equilibrium

[COCl2] = 0.3166 - x

[CO] = 0 + x

[Cl2] = 0 + x

c)

substitute in Kc

K = [CO][Cl2]/[COCl2]

2.2*10^-10 = x*x/(0.3166 -x)

solve for x

x = 8.345*10^-6

now..

[COCl2] = 0.3166 - 8.345*10^-6 = 0.316591 M

[CO] = 0 + 8.345*10^-6

[Cl2] = 0 + 8.345*10^-6

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