230_Sp10_L18_post - CHEM 230 Sp10 Lecture 18 Chapter 9: 1)...

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Unformatted text preview: CHEM 230 Sp10 Lecture 18 Chapter 9: 1) Temperature and K 2) Stress and Response Methods of Evaluation Hidden Stresses Session ID 230 Demos 2NO 2 (g) N 2 O 4 (g) H = -24.02 KJ/mol (red-brown) (colorless) CoCl 4 2- (aq) Co 2+ (aq) + 4Cl 1- (aq) 2 NO 2 (g) N 2 O 4 (g) H = -24.02 KJ/mol (red-brown) (colorless) cold: colorless higher K warm: dark red-brown lower K I 2(s) I 2(g) > Cold: more solid lower K Warm: purple gas higher K Temperature Effects For an endothermic reaction ( H > 0 ): Temperature increases favor the products ( K larger) Temperature decreases favor the reactants ( K smaller) For an exothermic reaction ( H < 0 ): Temperature increases favor the reactants ( K smaller) Temperature decreases favor the products ( K larger) For enthalpically neutral reactions ( H = 0 ): Temperature has no effect K = e "# H o RT e # S o R " G rxn o = # RT ln K = " H o # T " S o ln K = # " H o # T " S o ( ) RT = #" H o RT + " S o R Temperature and K K is constant at a particular temperature, but the absolute value does change with temperature For a system at equilibrium, the effect of temperature is related only to the enthalpy vant Hoff equation ln K 2 K 1 = " H rxn o R 1 T 1 # 1 T 2 $ % & ( ) Note: For reactions involving gases, use K (not K c ) Derived p 384 2NO (g) N 2(g) + O 2(g) K c(298 K) = 2.2 x10 30 = K p NO N 2 O 2 Rxn H o kJ/mol 90.25-180.5 S o J/(Kmol) 210.76 191.61 205.14-24.77 G o kJ/mol 86.55-173.1 Find K at 273K K @ 273 K Make sure to use J for all units Method 1: G = -RTlnK Method 2: ln K 2 K 1 = " H rxn o R 1 T 1 #...
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This note was uploaded on 01/24/2012 for the course PCHEM 230 taught by Professor Gottfried during the Fall '11 term at University of Michigan.

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230_Sp10_L18_post - CHEM 230 Sp10 Lecture 18 Chapter 9: 1)...

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