Entropy Tutorial

Example 20-9B (page 809)For the reaction2 NO(g) + Cl 2 (g) W 2 NOCl(g)∆GE = -40.9 kJ mol -1 , ∆HE = -77.1 kJ mol -1∆SE = -121.3 J mol -1 K -1 (all values correspond to 25EC)a) What is the value of K eq at 25EC?∆GE = -RT ln K eqIn K eq = -∆GE/RT= -(40.9 X 10 3 J mol -1 )/{(8.3145 J mol -1 K)(298.15 K)}= 16.5K eq = exp(16.5) = 1.47 X 10 7b) What is the value of K eq at 75EC?Method 1∆GE = ∆HE - T∆SE= (-77.1 kJ mol -1 ) - (348.15 K)(-121.3 X 10 -3 kJ mol -1 K -1 )= -34.9 kJ mol -1∆GE = -RT ln K eqIn K eq = -∆GE/RT= -(34.9 X 10 3 J mol -1 )/{(8.3145 J mol -1 K)(298.15 K)}= 12.06K eq = exp(12.06) = 1.7 X 10 5

Method 2o⎛K ⎞2∆H ⎛ 1 1 ⎞In⎜ ⎟ = ⎜ − ⎟K R T T⎝ 1 ⎠ ⎝ 1 2 ⎠3Keq77.1X10 1 1= −7In ⎛ ⎞ − ⎛ ⎞⎜ ⎟ 1.47X10 8.3145 ⎜ 298.15 348.15 ⎟⎝ ⎠⎝⎠7InKeq=− 4.467+ In(1.47X10 ) = 12.03K = exp(12.03) = 1.7X10eq5