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Chem 300 - Ch 26/#1 Today’s To Do List

Chem 300 - Ch 26/#1 Today’s To Do List. Start Chapter 26: Chem Equil: Extent of reaction (ξ) Deriva. of Δ r G o (T) = -RT ln K P (T) K P and T. Extent of Reaction. For general gas-phase reaction:  A A(g) +  B B (g)   Y Y (g) +  Z Z(g ) Extent of reaction (  )

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Chem 300 - Ch 26/#1 Today’s To Do List

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  1. Chem 300 - Ch 26/#1 Today’s To Do List • Start Chapter 26: Chem Equil: • Extent of reaction (ξ) • Deriva. of ΔrGo(T) = -RT ln KP(T) • KP and T

  2. Extent of Reaction • For general gas-phase reaction: • AA(g) + BB(g) YY(g) + ZZ(g) • Extent of reaction () • nA = A0 - A  • nY = Y0 + Y • As reaction proceeds: • dnA = - A d dnY = + Y d

  3. Enter Gibbs • For multicomponent system: dG = -S dT + V dP +(G/nA)dnA + (G/nB)dnB +(G/nY)dnY+(G/nZ)dnZ • dG = -SdT +VdP +A dnA+B dnB+Y dnY+ Y dnZ • At const T & P: • dG = - A Ad - B Bd + Y Yd + Z Zd • = (- A A- B B+ Y Y+ Z Z )d • (dG/ d )T,P = rG = Y Y+ Z Z- A A- B B

  4. The Reaction Quotient • Recall that • j(T,P) = oj(T) + RT ln(Pj/Po) • Subst in rG • rG = Y oY+ Z oZ- A oA- B oB + RT[Y ln(PY/Po) + Z ln(PZ/Po)- A ln(PA/Po)- B ln(PB/Po)] • rG = rGo + RTln Q • Q = Reaction Quotient = (PY PZ)/(PA PB )

  5. When at Equilibrium • (dG/ d )T,P = rG = 0 • 0 = rGo + RT ln Qeq • rGo = - RT ln Qeq = - RT ln[(PY PZ)/(PA PB )]eqrGo = - RT ln KP(T) • KP(T) = Thermo equilibr. Constant • Truly a constant at a given T

  6. In Terms of concentration • For a gas-phase reaction @ equilibr: • P = (n/V)RT = cRT • Subst. Into KP • KP=(cYYcZZ )/(cAAcBB )](RT)(Y+Z-A- B) • KP = KC (RT)(Y+Z-A- B)

  7. Example • NH3 = 3/2 H2 + ½ N2 • At 298 K KP = 1.36 x 10-3 • A = 1 Y = 3/2 Z = ½ • R = 0.0831 L-bar/mol-K • KP = KC (RT)1 • KC = 1.36 x 10-3/24.79 = 5.49 x 10-5

  8. Fract. of PCl3(g) dissoc. at equil. vs. Ptot at 473 K

  9. Next Time • Calculating KP from ΔrGo • When ΔrG is a minimum • ΔrG vs ΔrGo

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