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Some C6H5CH2OH is allowed to dissociate into C6H5CHO and H2 at 523 K. At equilibrium, [C6H5CH2OH]...

Some C6H5CH2OH is allowed to dissociate into C6H5CHO and H2 at 523 K. At equilibrium, [C6H5CH2OH] = 0.212 M, and [C6H5CHO] = [H2] = 5.24×10-2 M. Additional C6H5CHO is added so that [C6H5CHO]new = 9.85×10-2 M and the system is allowed to once again reach equilibrium.

  • C6H5CH2OH(g) 7e50f1f7-d29b-4777-b1ee-365a063b4be8.gifC6H5CHO(g) + H2(g) K = 1.30×10-2 at 523 K


(a) In which direction will the reaction proceed to reach equilibrium?

(b) What are the new concentrations of reactants and products after the system reaches equilibrium?

[C6H5CH2OH] = M
[C6H5CHO] = M
[H2] = M
0 0
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Answer #1

The given reaction is

CHECH,OH) = C6H5CHO) + H2(g)

The starting equilibrium concentrations are

C6H3CH2OH = 0.212 M

CH3CHO= H2 = 5.24 x 10-2 M

Note that the equilibrium constant K is 1.3 x 10-2 at 523 K .

(a).

Now, the equilibrium is disturbed by addition of additional C.HSCHO. Note that it is a product in our reaction scheme.

Hence, according to Le Chatelier principle, the equilibrium shifts to counteract the change created by a disturbance to the equilibrium.

Now, since we have added some product to the starting equilibrium, the reaction will remove some products and form reactants to reach the new equilibrium.

This can be mathematically shown by calculating the reaction quotient Q after disturbance.

C.HSCHOOL = 9.85 x 10-2 M

H2 = 5.24 x 10-2 M

C6H3CH2OH = 0.212 M

Hence, the reaction quotient Q is given by

[C.HSCHOnew X (H) (C6H3CH2OH 9.85 x 10-2 X 5.24 x 10-2 - = 2.43 x 10-2 0.212

Note that the reaction quotient Q > K.

Hence, there is more product than the equilibrium amount after disturbance by addition of a product. Hence, the reaction will proceed towards left (i.e. towards reactants) to attain equilibrium.

(b).

To calculate the new concentrations of reactants and products after equilibrium is reestablished, we will create the following ICE table

C.HSCHOH C.HSCHO H_2
Initial, M 0.212 9.85 \times 10^{-2} 5.24 x 10-2
Change, M +x -x -x
Equilibrium, M 0.212+x 9.85 \times 10^{-2}-x 5.24 \times 10^{-2}-x

Note that for x M of products consumed, we will get x M of reactant formed.

Hence, from the ICE table, we can write the following expression of K

K = H2CH3CHO C6H5CH OH (5.24 x 10-2-2) (9.85 x 10-2-2) K= 0.212 +2 20.0163 M = 1.63 x 10-2 M = 1.30 x 10-2

Hence, the new equilibrium concentrations can be calculated as follows

[C_6H_5CH_2OH] = 0.212 +x = 0.212 \ M + 1.63 \times 10^{-2} \ M \approx {\color{Red} 0.228 \ M}

ICH,CHO = 9.85 x 10-2 M - 1.63 x 10-2 M 8 .22 x 10-?M

[H_2] = 5.24 \times 10^{-2} \ M - 1.63 \times 10^{-2} \ M \approx {\color{Red} 3.61 \times 10^{-2} \ M}

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Answer #2
Pcl5 (g) = pcl3 (g)+cl2 (g)
source: Chemistry
answered by: William
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