Consider the following equilibrium: 2NOCl (g) 2NO (g) + Cl_2 (g) Delta G degree

Question

Consider the following equilibrium: 2NOCl (g) 2NO (g) + Cl_2 (g) Delta G degree = 41. kJ Now suppose a reaction vessel is filled with 9.72 atm of nitrosyl chloride (NOCl) and 1.80 atm of nitrogen monoxide (NO) at 883. degree C. Answer the following questions about this system: Under these conditions, will the pressure of NOCl tend to rise or fall? rise fall Is it possible to reverse this tendency by adding Cl_2? In other words, if you said the pressure of NOCl will tend to rise, can that be changed to a tendency to fall by adding Cl_2? Similarly, if you said the pressure of NOCl will tend to fall, can that be changed to a tendency to rise by adding Cl_2? yes no If you said the tendency can be reversed in the second question, calculate the minimum pressure of Cl_2, needed to reverse it. Round your answer to 2 significant digits. atm

Explanation / Answer

1)

Since initially p(Cl2) is 0, the reaction is going to move to product side

So, the pressure of NOCl will fall

2)

Yes, it is possible to reverse the tendency

when we make Qp greater than Kp by adding some extra Cl2

3)

G = 41 KJ = 41000 J

T = 883.0 oC =(883.0+273)K = 1156.0 K

use:

G = -R*T*ln Kc

41000 = - 8.314*1156.0* ln(Kc)

ln Kc = -4.266

Kc = 1.404*10^-2

now calculate Kp,

n = number of gaseous molecule in product - number of gaseous molecule in reactant

n = 1

Kp= Kc (RT)^ n

Kp = 0.01404*(0.0821*1156.0)^(1)

Kp = 1.333

Qp > Kp

p(Cl2)*p(NO)^2 / p(NOCl)^2 > 1.333

p(Cl2) * (1.80)^2 / (9.72)^2 > 1.333

p(Cl2) = 39 atm

Answer: 39 atm

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