{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chap-15-2011W-a

# Varies with temperature p636 fig 157 k value 10 5

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: nt. •varies with temperature p636: Fig. 15.7 K value 10-5 Reaction Hardly occurs 10-3 1 103 Reactants & products Co-exist in equilibrium 105 Goes to completion Questions involving Equilibrium Calculations K = 3.3×10-13 for AgBr(s) -16 K = 1.5 ×10 for AgI(s) Ag+ (aq) + Br - (aq) + - Ag (aq) + Cl (aq) (a) If [Ag+] = 5.7×10-7 M, what is [Br-] at equilibrium? K = [ Ag + c ][ Br − c ] K = [ Ag + c ][ Br − c ] = (5.7 ×10−7 M /1M ) ⋅ [ Br − /1M ] = 3.3 ×10−13 − −7 [ Br ] = 5.8 ×10 M (b) Is AgI(s) more soluble than AgBr(s)? Explain. No, it is less soluble …. K (AgI)is smaller! Question: Given that when K = 1 for this reaction N2(g) + 3H2(g) ↔ 2 NH3(g) H = + 92 kJ there are [N2] = [H2] = [NH3] = 1.0 mol/L, If the reaction vessel is suddenly decrease to ½ of its existing size, what is Q at that “instant”? 2 2 ([ NH 3 ] c ] K= K= 3 ([ N 2 ] c ) ⋅ ([ H 2 ] c ) (1mol / L /1mol / L) (1mol / L /1mol / L) ⋅ (1mol / L /1mol / L) 3 =1 Concentration = 1.0 mol ÷ 0.5 L = 2.0 M 2 Q= (2mol / L /1mol / L) (2mol / L /1mo...
View Full Document

{[ snackBarMessage ]}