{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

122ch15d

# 122ch15d - 40 15.67 Kp = Kc(RT)n Kc = Kp(RT)n...

This preview shows pages 1–10. Sign up to view the full content.

40 15.67) K p = K c (RT) ) n K c = K p (RT) ) n = (4.3 3 2) {(0.0821 L C atm/mol C K)(1100 K)} ( 1) = 0.047 9 68 = 0.0480

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

View Full Document
41 15.70)
42 15.71)

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

View Full Document
43 15.72)
44 15.74) Mathematical LeChatelier’s Principle Application

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

View Full Document
45 15.76)
46 15.77) 15.79)

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

View Full Document
47 15.80)
48 15.81) Both the forward and reverse reactions in the equilibrium A º B are elementary (single-step) reactions. Assume the only effect a catalyst has on the reaction is to lower the activation energies of the forward and reverse directions by the same amount (i.e. it lowers the “hill” in the reaction pathway by the same amount). Prove that the equilibrium constant for the catalyzed reaction is the same as for the uncatalyzed reaction. The catalyst only speeds up the reaction and does not effect the position of equilibrium. The amount of reactants and products at equilibrium is the same with or without a catalyst. Remember, for a reaction in which both the forward and reverse reactions are elementary reactions you can show the equilibrium constant is related to the rate

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 / 11

122ch15d - 40 15.67 Kp = Kc(RT)n Kc = Kp(RT)n...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online