Consider the dissociation of nitrosyl cholride: NOCl(g) NO_(g) + 1/2 Cl_2 (g) If

Question

Consider the dissociation of nitrosyl cholride: NOCl(g) NO_(g) + 1/2 Cl_2 (g) If 1.000 mole of NOCl gas is initially added to a flask held at 1.000 bar, a. Derive an expression for K_p in terms of zeta. B. Calculate (Using table 3) Delta_f H degree, Delta_g G degree, Delta_g S degree, Delta_g C_p, and K_p for the reaction at 298.15 K. c. Solve for zeta at equilibrium and 298.15 K. D. Using the Gibbs-Helmholtz equation and the van't Hoff equation, calculate Delta_f G degree, K_p, and zeta at 600.0 K. Table 3: Delta_g H degree, Delta_g G degree, Delta_g S degree, and C_p of reactants and products.

Explanation / Answer

The reaction is NoCl<-> NO +1/2Cl2

Initially , there is 1 mole of NOCl

Inital pressure of NOCl= 1bar

Let x= drop in pressure of NOCl to reach equilibrium

So at equilibrium , partial pressures : NOCl = 1-x, NO= x and Cl2= x/2

KP= [PNO] [PCl2]0.5/ [PNOCl], P indicates the partial pressue of NoCl

Kp = x*(x/2)0.5/(1-x)

deltaH0= enthalpy of products- enthalpy of reactants = 91.3*1- (51.7) = 39.6 Kj/mol

deltaS= 210.8+0.5*223.1- (261.7) =60.65 J/mol.K

deltaGO= 87.6- 66.1= 21.5 KJ/mol

deltaG= -RT lnK

lnK= -21.5*1000/(298.15*8.314)

K= 0.000171= x*(x/2)0.5/ (1-x) = 0.707* x 1.5/ (1-x)

Assuming   x<<1, 0.707*x0.5= 0.000171, x= 0.01555

At 600K

deltaG= deltaH- TdeltaS = 39.6*1000- 600*60.65 =3210 J/mole= 3.210 KJ/mole

from Van’t Hoff equation ln (K2/K1)= (deltaH/R)*(1/T1-1/T2)

K2 = equilibrium constant at 600K and K1 = equilibrium constant at 298.15

Ln(K2/0.000171)= (39.6*1000/8.314)*(1/298.15-1/600)=8.04

K2/0.000171= 3102.613

K2= 3102.613*0.000171= 0.53 = 0.707* x 1.5/ (1-x)

When solved using excel, x= 0.51

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