Be sure to answer all parts.Using the data in the table, calculate Kp and AGfor this reaction at 25°C.PC15(g) = PC13(g) + Cl2(g)Initial PartialAGSubstancePressure(kJ/mol) (atm)C2(8)0.353PCl3(g)-2860.216PCI5(g)-3250.00290Кр-X 10AG =kJ/mol

For a general reaction aA + bB → cc + dD ∆G°rxn =( Σv G°f )product - ( Σ vG°f )reactant ∆G°rxn = [c (∆G°f )c + d (∆G °f )d] - [a(∆G °f )a + b(∆G °f )b] At equilibrium, ∆Grxn = 0.0 kJ/mole and Q = Kp ∆G°rxn = - 2.303RTlog Kp and , ∆Grxn = ∆G°rxn + 2.303RTlog Q For the reaction, PCl5(g) ⇌ PCl3(g) + PCl3 (g) ∆G°rxn = [ (∆G°f ) PCl3 + d (∆G °f ) Cl2] - [a(∆G °f ) PCl5 ] ∆G°rxn = [ -286 kJ/mol + 0 kJ/mol ] - [-325 kJ/mol ] ∆G°rxn =39 kJ/mol At equilibrium, ∆Grxn = 0.0 kJ/mole and Q = Ka ∆G°rxn = - 2.303RTlog Kp = - 2.303 x 8.314 x 298 logKp 39 kJ/mol =-5705.85 log Kp Kp = 10(-39000/5705.85) =10-6.835 = 1.46 x 10-7 ∆Grxn = ∆G°rxn + 2.303RTlog Q Q = P PCl3 x P Cl2 / P PCl5 From the given data, Q = 0.216 x 0.353 / 0.00290 = 26.29 So, ∆Grxn = ∆G°rxn + 2.303RTlog Q ∆Grxn = 39000 J/mol + 2.303 x 8.314 x 298 log 26.29 ∆Grxn = 47101.34 J/mol ∆Grxn = 47.1 kJ/mol

Answered: Be sure to answer all parts. Using the…

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Chemistry

Be sure to answer all parts. Using the data in the table, calculate Kp and AGfor this reaction at 25°C. PCI5(g) = PC138) + C2(g)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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