5. The reaction for the formation of nitrosyl chloride, 2 No(g) C2(g)2 NOCI(g),

Question

5. The reaction for the formation of nitrosyl chloride, 2 No(g) C2(g)2 NOCI(g), was studied at 25°C. The pressures at equilibrium were found to be PNOCI-1.2 atm, pNo 5.0 x 10-2 atm, and PCI.-3.0 × 10-1 atm. Standard pressure is 1.0 bar-0.986923 atm (a) Caleulate the vale of K, for this reaction at 25 c (b) If 1.0atm of NOCI(g) was placed into an evacnated container, would a reaction occur? (c) If 1.0atm each of NOCl(g). NO(g). and Clz(g) were placed into au evacuated container, would a reaction occur? If so, what direction would the reaction proceed? (d) If 1.0 atiu of NOCI(g), 5.0 × i0-3 at m of NO(g), and 3.0 × 10-1 attn of C12(g) were placed into an evacuated container, would a reaction occur? If so, what direction would the reaction proceed? Show all calculations and explain each answer

Explanation / Answer

a)

Kp = P-products / P-reactants

Kp = NOCl^2 /(NO^2)(Cl2)

Kp = (1.2^2)/(((5*10^-2)^2)(3*10^-1))

Kp = 1920

b)

yes, NOCl is a product, so it will form more reactants NO, Cl2 must increase

c)

yes it will be reaction..

Q= ((1.2+1)^2)/(((1+5*10^-2)^2)(1+3*10^-1))

Q = 3.3769

since Q << Kp

then, there will be more reactants than products, expect product formation, i.e. right shift

d)

Q = NOCl^2 /(NO^2)(Cl2)

Q = (1^2) /((5*10^-3)^2)(3*10^-1))

Q = 12000

since Q > Kp

expect more products than reactanants, therefore, the shift must go to produce even more reactant, i.e. No and Cl2

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