{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture 19sf

# Lecture 19sf - Temperature Dependence of K o o o G = H T S...

This preview shows pages 1–4. Sign up to view the full content.

Temperature Dependence of K G o = H o - T S o = -RT ln K Dividing both sides by -RT and rearranging: ln K = - Τ ∆Η ο R + R S ο Compare: y = mx + b (straight line equation) where y = ln K and x = Plot ln K vs and get a straight line. The slope of the line = - R ο ∆Η and the y-intercept = R S ο This slope will be positive or negative depending on the sign of H. Suppose the plot of ln K vs gives a positive slope. Since the slope - R ο ∆Η , a positive slope implies H is negative (exothermic reaction). A positive slope also means: As ln K increases, increases As ln K increases, T decreases As K increases, T decreases As K decreases, T increases This is consistent with LeChatelier’s principle, which implies that for an exothermic reaction, a lower temperature will shift the reaction to the right leading to a higher value of K, or conversely, a higher temperature will shift the reaction to the left, leading to a lower value of K. To compare two temperatures and two equilibrium constants:

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

View Full Document
ln K 2 = - 2 R Τ ∆Η ο + R S ο ln K 1 = - 1 R Τ ∆Η ο + R S ο Subtracting the second equation from the first: ln K 2 – ln K 1 = - R ο ∆Η ( 2 T 1 - 1 T 1 ) This is equivalent to: ln 1 2 K K = R ο ∆Η ( 1 T 1 - 2 T 1 ) or ln 1 2 K K = R ο ∆Η ( 2 1 1 2 T T T T - ) N 2 (g) + 3H 2 (g) 2NH 3 (g) K = 6.0 x 10 5 at 298.15 K. What is value of K at 600 K? For this reaction, H o = -92.22 kJ G o = -32.96 kJ S o = -198.7 J/K ln 1 2 K K = R ο ∆Η ( 1 T 1 - 2 T 1 ) ln 5 10 x 0 . 6 x = 314 . 8 220 , 92 - ( ) 15 . 298 )( 600 ( 15 . 298 600 - ) Solving: x = 4.46 x 10 -3 Alternate method:
G o = -RT ln K G o = -32.98 kJ at 298 K. But this value will change at other temperatures. G o = H - T S o AT 600 K: G o = -92.220 -600( 1000 7 . 198 - ) = +27.0 kJ Note that reaction is spontaneous in the reverse direction at this high temperature. +27.0 = - ln K K = 4.46 x 10 -3 Note that: K = 6.0 x 10 5 at 298 K K = 4.46 x 10 -3 at 600 K K is lower at higher temperature which is expected for an exothermic reaction ( H < 0).

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

Lecture 19sf - Temperature Dependence of K o o o G = H T S...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online