Question

Q.26

Estimate the value of the equilibrium constant at 620 K for each of the following reactions. ΔG∘f for BrCl(g) is −1.0 kJ/mol. The standard molar entropy, S∘, for BrCl(g) is 240.0 J/mol⋅K.

Part A

2NO2(g)⇌N2O4(g). ΔH∘f for N2O4(g) is 9.16 kJ/mol.

Express your answer using three significant figures.

K =

Part B

Br2(g)+Cl2(g)⇌2BrCl(g) .ΔH∘f for BrCl(g) is 14.6 kJ/mol.

Express your answer using three significant figures.

K =

Answer 1

Sol.

Part A :  

2NO2(g) <---> N2O4(g)

So , Standard change in enthalpy = deltaH°  

= deltaHf°(N2O4 , g) - 2 × deltaHf°(NO2 , g )

= 9.16 - 2 × 33.18 = - 57.2 KJ

Standard change in entropy = deltaS°

= deltaSf°(N2O4 , g ) - 2 × deltaSf°(NO2 , g )

= 304.29 - 2 × 240.06

= - 175.83 J/K

= - 0.17583 KJ/K

So , Standard change in free energy

= deltaG° = deltaH° - T × deltaS°

where T = Temperature = 620 K

So , deltaG° = - 57.2 - 620 × ( - 0.17583 ) = 51.8146 KJ

Also , deltaG° = - R × T × ln(K)

where R = Gas constant = 0.008314 KJ / K mol

T = 620 K

K = equilibrium constant  

So , 51.8146 = - 0.008314 × 620 × ln(K)

ln(K) = - 10.0519

K = e^-10.0519 = 4.31 × 10^-5

Part B :

Br2(g) + Cl2(g) ---> 2BrCl(g)

deltaH° = 2 × deltaHf°(BrCl , g ) - deltaHf°(Br2 , g ) - deltaHf°(Cl2 , g )

= 2 × 14.6 - 30.907 - 0 = 1.707 KJ  

deltaS° = 2 × deltaSf°(BrCl , g ) - deltaSf°(Br2 , g) - deltaSf°(Cl2 , g )  

= 2 × 240.0 - 245.463 - 223.066

= 11.471 J / K

= 0.011471 KJ / K

So , deltaG° = 1.707 - 620 × 0.011471 = - 5.40502 KJ

And , ln(K) = - 5.40502 / ( - 0.008314 × 620 ) = 1.0485

K = e^1.0485 = 2.8533