Question

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) C₆H₅CHO(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

Answer 1

The given reaction is

$C_6H_5CH_2OH_{(g)} \rightleftharpoons C_6H_5CHO_{(g)} + H_2_{(g)}$

The starting equilibrium concentrations are

$\left [C_6H_5CH_2OH_{(g)} \right ] = 0.212 \ M$

$\left [C_6H_5CHO_{(g)} \right ] = [H_2] = 5.24 \times 10^{-2} \ M$

Note that the equilibrium constant K is $1.3 \times 10^{-2} \ at \ 523 \ K$ .

(a).

Now, the equilibrium is disturbed by addition of additional $C_6H_5CHO$. 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.

$\left [C_6H_5CHO_{(g)} \right ]_{new} = 9.85 \times 10^{-2} \ M$

$[H_2] = 5.24 \times 10^{-2} \ M$

$\left [C_6H_5CH_2OH_{(g)} \right ] = 0.212 \ M$

Hence, the reaction quotient Q is given by

$Q = \frac{[C_6H_5CHO]_{new} \times [H_2]}{[C_6H_5CH_2OH]} = \frac{9.85 \times 10^{-2} \times 5.24 \times 10^{-2}}{0.212} = 2.43 \times 10^{-2}$

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

	$C_6H_5CH_2OH$	$C_6H_5CHO$	$H_2$
Initial, M	0.212	$9.85 \times 10^{-2}$	$5.24 \times 10^{-2}$
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

$K = \frac{[H_2][C_6H_5CHO]}{[C_6H_5CH_2OH]} \\ \Rightarrow K= \frac{(5.24 \times 10^{-2}-x) \times (9.85 \times 10^{-2}-x)}{0.212+x} = 1.30 \times 10^{-2} \\ \\ \Rightarrow x \approx 0.0163 \ M = 1.63 \times 10^{-2} \ M$

Hence, the new equilibrium concentrations can be calculated as follows

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

$[C_6H_5CHO] = 9.85 \times 10^{-2} \ M - 1.63 \times 10^{-2} \ M \approx {\color{Red} 8.22 \times 10^{-2} \ M}$

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

Answer 2

Pcl5 (g) = pcl3 (g)+cl2 (g)