HW Set 18b-Google Chrome Secure https//session.masteringchemistry.com/myct,itemV

ID: 592492 • Letter: H

Question

HW Set 18b-Google Chrome Secure https//session.masteringchemistry.com/myct,itemView mrseLUR Execse 18 78 Enhanced . with Feedback ssignmentProblemilD-939266048offset next Exercise 18.78 Enhanced with Feedback Part A Estimate the value of the equilbrium constant at 650 K for each of the following reactions AGi for BrCl() is-10 Express your answer using three significant figures /mol The standard molar entropy 240 0 J/mol K for Bra(g) is You may want to reference (LD pages 873-878) Section 18 10 while completing this problem Incorrect; Try Again;5 attempts remaining Calculate the change in Gibb's free enargy AG,using the equton where 1,isthe-tha py change tor reaction, ,"istheenhopy change for th. reden and rhth. Kakibmper at "Thrus. the calculated value for AG and the equation ehere R is the gas constant (8314 J/K-mol and T is the Kevi temperature to sove for the equilrium constant K Part Brlt) ·Cli (g) 2nrCl(r) , //, tor Elecug)s 14 6 LI ind Express your answer using three significant figres Type here to search

Explanation / Answer

Part A

The standard enthalpy and entropy of formation values are required (taken from internet sources)

H0f (kJ/mol)

S0f (J/mol.K)

NO2 (g)

33.2

240.0

N2O4 (g)

9.16

304.2

Consider standard conditions, i.e, T = 25°C = (25 + 273) K = 298 K

H0rxn = [(1 mole)*H0f (N2O4)] – [(2 mole)*H0f (NO2)] = [(1 mole)*(9.16 kJ/mol)] – [(2 mole)*(33.2 kJ/mol)] = -57.24 kJ.

S0rxn = [(1 mole)*S0f (N2O4)] – [(2 mole)*S0f (NO2)] = [(1 mole)*(304.2 J/mol.K)] – [(2 mole)*(240.0 J/mol.K)] = -175.8 J/K.

G0rxn = H0rxn – T*S0rxn = -(57.24 kJ) – (298 K)*(-175.8 J/K) = -57.24 kJ – (-52388.4 J)*(1 kJ/1000 J) = -57.24 kJ – (-52.3884 kJ) = -57.24 kJ + 52.3884 kJ = -4.8516 kJ = -(4.8516 kJ)*(1000 J/1 kJ) = -4851.6 J.

Use the relation G0rxn = -R*T*ln K to find the equilibrium constant, K.

-4851.6 J = -(8.314 J/mol.K)*(298 K)*ln K

====> 4851.6 J = (2477.572 J)*ln K

====> ln K = 1.9582

====> K = exp^(1.9582) = 7.0865 7.086 (ans).