Question

Consider the following equilibrium: 2NOCI (8) - 2NO($)+C12(e) AGO 41. kJ Now suppose a reaction vessel is filled with 8.25 atm of nitrosyl chloride (NOCI) and 1.95 atm of nitrogen monoxide (NO) at 800. "C. Answer the following questions about this system: rise Under these conditions, will the pressure of NOCI tend to rise or fall? XS ? Is it possible to reverse this tendency by adding Cl? In other words, if you said the pressure of NOCI will tend to rise, can that be changed to a tendency to fall by adding C1,7 Similarly, If you said the pressure of NOCI will tend to fall, can that be changed to a tendency to rise by adding C, If you said the tendency can be reversed in the second question, calculate the minimum pressure of Ci, needed to reverse it Round your answer to 2 significant digits.

Answer 1

We have equilibrium:

2 NOCl (g) $\rightleftharpoons$ 2 NO (g) + Cl₂ (g)   $\Delta$G = 41 kJ

In reaction vessel : P_NOCl = 8.25 atm P_NO = 1.95 atm (at 800 C = 1073 K)

Kp = ( P_NO )2 ( P_Cl2 ) / (P_NOCl) = Kc (RT)

$\Delta$G = 41 kJ , we have  $\Delta$G = -RT ln Kc

so, lnKc = -$\Delta$G /RT = -4.6

or Kc = 0.01

and Kp = 89.2

1. Under these conditions :

Reaction quotient , Q = ( P_NO )2 ( P_Cl2 ) / (P_NOCl) = ( 1.95)2 / (8.25 ) = 0.46

It is less than Kp= ​​​​​​​ 89.2 ; Kp > Q

Since, If Q < K , the reaction will proceed in the direction of the products (forward reaction).

So, reaction will move in forward direction thus fall in pressure of NOCl.

2. If we add sufficient amount of   Cl₂ that Kp < Q , than this tendency can be reversed.

Since, Q > K , the reaction will proceed in the direction of reactants (reverse reaction).

3. Minimum pressure of Cl₂ required to reverse the tendency,

it will be at , Kp = Q

Q = ( P_NO )2 ( P_Cl2 ) / (P_NOCl) = ( 1.95)2 ( P_Cl2 )/ (8.25 ) = 89.2

0.46* ( P_Cl2 ) = 89.2

or ,  P_Cl2 = 193.55 atm ~ 190 ( 2 significant digits)