{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chem102-Chap_14

# 92 4 k2 33x10 k3 4 ch 4 h 2 o cs2 h2 k 2 coh

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: centration of a gas = = P( ) V RT · Kp is the equilibrium constant for a gaseous reaction expressed in terms of partial pressures. · Kp has a value different fro m Kc Kp = Kc(RT)Dn Dn = sum of coefficients sum of coefficients o f – o f gaseous products gaseous reactants CO (g) + 3 H2 (g) Example 1: K p = ¾¾ ® ¬¾ ¾ PCO P 23 H H2O(g) Kc = 3.92 PCH4 P 2O H CH4(g) + (at 1200K) Dn = (1+1) – (1+3) = – 2 Kp = Kc(RT)Dn = 3.92 [(0.0821) (1200)]-2 = 4.04 x 10-4 Example 2: The equilibrium constant Kc equals 10.5 for the fo llowing reaction at 227 °C. CO(g) + ¾¾ ® 2 H2 (g) ¬¾ CH3OH (g) ¾ What is the value o f Kp at this temperature? Dn = Kp = Kc(RT)Dn = 9 Chemistry 102 Chapter 14 OTHER EQUILIBRIUM CONSTANT · If a given chemical equat ion can be obtained by taking the sum of other equations, the Equilibrium Constant for the overall equat ion equals the product of the equilibrium constants of the other equations. Koverall = K1K2… Example 1: The fo llowing equilibria occur at 1200K CO(g) + ¾¾ ® 3 H2 (g) ¬¾ CH4 (g) ¾ + H2O(g) CH4(g) + ¾¾ ® 2 H2S (g) ¬¾ CS2 (g) ¾ + 4 H2 (g) CO(g) + ¾¾ ® 2H2S (g) ¬¾ ¾ K1 = CS2 (g) + H2O (g) + H2 (g) K1 = 3.92 4 K2 = 3....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online