Lecture 11

Lecture 11 - 2/1/10 Consider the following reaction: H2 (g)...

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

View Full Document Right Arrow Icon
2/1/10 1 8 Consider the following reaction: H 2 (g) + F 2 (g) ! 2HF (g) What is the expression for K? A) B) C) D) K = HF [ ] H 2 [ ] F 2 [ ] K = 2 HF [ ] F 2 [ ] H 2 [ ] K = HF [ ] 2 H 2 [ ] F 2 [ ] K = H 2 [ ] F 2 [ ] HF [ ] 2 Form: Products (RHS) Reactants (LHS) Exponents given by stoichiometric coefficients 1.A. Equilibrium Constants using Pressures of Gases For gases, use pressure instead of molarity (mol/L ) ** Given concentrations, use them to calculate K Given pressures, use them to calculate K All equilibrium constants K 9 H 2 (g) + F 2 (g) ! 2HF (g) K = P HF 2 P H 2 P F 2 Consider the previous reaction: K is unitless Equilibrium Constant: P = pressure in bar K is constant at a given temperature (even if pressures do not seem to cancel) (or atm since 1 bar ! 1 atm) (K c ) (K p ) Book uses: K c , K p , K a , K b , K w , K eq , K sp …. KP = P NO 2 2 PNO 2 PO 2 = KCRT ! 1 If, (for above reaction) Generally, K P = K c RT ( ) ! n gas Where " n gas = mols of gas product - mols of gas reactant For above reaction, " n gas = 2 - 3 = -1 K = K c is overall equilibrium constant for gases or solutions 1. Equilibrium Constants using Pressures of Gases (Section 17.3) Ideal Gas Law : PV = nRT P = pressure (atm) V = volume (L) n = mols R = 0.08206 L atm mol -1 K -1 T = Temperature (Kelvin) UNITS !! data sheet
Background image of page 1

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

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

Page1 / 3

Lecture 11 - 2/1/10 Consider the following reaction: H2 (g)...

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