Lecture 30 - e.g SiH4(g 2 H2O(g SiO2(s 4 H2(g If K at...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
3/27/10 1 e.g. SiH 4 (g) + 2 H 2 O(g) ! SiO 2 (s) + 4 H 2 (g) If K at 298.15 K is 8.81 ! 10 79 , calculate K at 1273 K Assume " H o not T-dependent 64 " H º = 4 " H º f ( H 2 (g) ) + " H º f ( SiO 2 (s) ) ! " H º f ( SiH 4 (g) ) ! 2 " H º f ( H 2 O(g) ) = ! 461.6 kJ = 4 mol(0kJ/mol)+( # 910.9kJ) # (34.3kJ) # 2mol( # 241.8kJ/mol) = ! 461.6 kJ/mol of rxn ln K 2 K 1 ! " # $ % = ’( H o R 1 T 2 1 T 1 ! " # $ % = ! ! 461.6 " 10 3 J / mol ( ) 8.314 J / molK 1 1273 K ! 1 298.15 K # $ % = ! 142.60 K 2 = K 1 e ! 142.60 = (8.81 ! 10 79 )e ! 142.60 = 1.03 ! 10 18 Use van’t Hoff equation: Need " H o 65 Reaction shifts to the left SiH 4 (g) + 2 H 2 O(g) ! SiO 2 (s) + 4 H 2 (g) Why has K decreased with increasing T ? Reaction is exothermic Less products and more reactants Increase T: equilibrium shifts to consume heat " H º = ! 461.6 kJ + heat = smaller K (Le Chatelier’s principle) 66 " G º = 4 " G º f ( H 2 (g) ) + " G º f ( SiO 2 (s) ) ! " G º f ( SiH 4 (g) ) ! 2 " G º f ( H
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

Lecture 30 - e.g SiH4(g 2 H2O(g SiO2(s 4 H2(g If K at...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online