Consider the following equilibrium: Now suppose a reaction vessel is filled with

Question

Consider the following equilibrium: Now suppose a reaction vessel is filled with 8.13 atm of nitrosyl chloride (NOCI) and 2.30 atm of nitrogen monoxide (NO) at 1168. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of NO tend to rise or fall? 10 fall Is it possible to reverse this tendency by adding Cl2? In other words, if you said the pressure of NO will tend to rise, can that be changed to a tendency to fall by adding CI 2? Similarly, if you said the pressure of NO will tend to fall, can that be changed to a tendency to rise by adding Cl2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of CI2 needed to reverse it. Round your answer to 2 significant digits. yes no atm

Explanation / Answer

To know this we have to find reaction quotient(Q)

Q = [NO]2[Cl2]/[NOCl]

Q = 0 here as Cl2 is = 0 initially

if Q < K reaction goes forward

NO pressure will rise

Yes its possible to reverse this tendency as we keep of increasing Cl2 in our reaction , Q will also keep on increasing and as soon as it crosses(more than) equilibrium constant K , reaction reverses

yes it has tendency to fall, when Cl2 is added

1168 oC = 1168 + 273 = 1441 K

Initial concentration of NOCl = n/V = P/RT = 8.13 atm / (0.08206 atm.L/mol.K x 1441K) =0.06875 M

NO = 2.3 atm / (0.08206 atm.L/mol.K x 1441K) = 0.01945 M

G = -RTlnK

41000J = - 8.314 x 1441K lnK

K = 0.0326389

2NOCl <-------> 2NO + Cl2

0.06875 M   0.01945 M 0 Initially

to reverse ,we should get Q=>K

K = [0.01945]2[Cl2] / [0.06875]2 = 0.0326389

[Cl2] = 0.4078 M

pressure = P/RT = 0.4078 mol/L

P = 0.4078 mol/L x 0.08206 atm.L/mol.K x 1441K = 48.22 atm

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