Given the following values for components of a system, at 298.15K and 1 atm: PCl

Question

Given the following values for components of a system, at 298.15K and 1 atm: PCl5, delta H = -374.9 kJ/mol, S = 364.6 J/mol K PCl3, delta H = -287 kJ/mol, S = 311.8 J/mol K Cl2, delta H = 0 kJ/ mol, S = 223 J/mol K
Calculate the values for Kp at 298.15K for PCl5 (g) <--> PCl3(g) + Cl2(g) Given the following values for components of a system, at 298.15K and 1 atm: PCl5, delta H = -374.9 kJ/mol, S = 364.6 J/mol K PCl3, delta H = -287 kJ/mol, S = 311.8 J/mol K Cl2, delta H = 0 kJ/ mol, S = 223 J/mol K
Calculate the values for Kp at 298.15K for PCl5 (g) <--> PCl3(g) + Cl2(g) PCl5, delta H = -374.9 kJ/mol, S = 364.6 J/mol K PCl3, delta H = -287 kJ/mol, S = 311.8 J/mol K Cl2, delta H = 0 kJ/ mol, S = 223 J/mol K
Calculate the values for Kp at 298.15K for PCl5 (g) <--> PCl3(g) + Cl2(g)

Explanation / Answer

We use delta H and delta S of the reaction to calculate delta G

Delta Hrxn = [1mol*delta H PCl3+1 mol* delta H Cl2]-[1 mol * delta H PCl5]

=[1 mol * -287 kJ/mol + 1 mol * 0 ]-[1 mol * -374.9 kJ/mol]

= -287 kJ – (-347.9 kJ)

= 60.9 kJ

= 60900 J

Delta Srxn = [1mol*delta S PCl3+1 mol* delta S Cl2]-[1 mol * delta S PCl5]

= (1mol * 311.8 J per K per mol + 1 mol * 223 J per K per mol )

       -(364.6 J per K per mol )

= 170.2 J per K

We know

Delta G = Delta H – T delta S

T = 298.15 K

Delta G = 60900 J – 298.15 K * 170.2 J per K

= 10154.87 J

We get kp for this reaction by using following equation

Delta G = -RTlnKp

Here R = 8.314 J per K per mol

10154.87 J = - ( 8.314 J per K per mol * 298.15 K * ln K )

-4.097 = ln K

Lets take exp of both side

0.01663 = K

Kp of this reaction = 0.01663

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