Intermolecular forces/Liquid/Gas phase/Kinetics

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20 Terms

1
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Hydrogen bonding (strongest)

F-H, O-H, or N-H bond required in a pure substance

F, O, or N to hydrogen bond with H2O

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dipole-dipole forces (2nd strongest)

intermolecular force for polar molecules

3
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London dispersion forces (Van der Waals)

A temporary or transient dipole

All molecules have these and the greater the weight and surface area, the greater the London dispersion forces

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Higher intermolecular forces lead to —-

Higher boiling point

higher heat of vaporization

higher viscosity

higher surface tension

lower vapor pressure

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Ideal Gas Law

PV=nRT

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Boyle’s Law

P = 1/V

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Charle’s Law

V = T

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Avogadro’s Law

V = n

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combined gas law

PV/nT = PV/nT

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catalyst

Speeds up a reaction by lowering the activation energy, by providing an alternate mechanism for the rxn to occur

  • NOT consumed in a rxn

  • does NOT shift equilibrium

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kinetic product

lower activation energy

favored at low temperatures

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thermodynamic product

Overall lower energy product

favored at higher temperatures

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reaction order zero

(A) = (A)0 - kt

t1/2 = (A)0/2k — the less reactant the shorter it will take

<p>(A) = (A)<sub>0</sub> - kt</p><p>t1/2 = (A)<sub>0</sub>/2k — the less reactant the shorter it will take </p>
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1st order rxn

ln(A) = ln(A)0 -kt

t1/2= 0.693/k — does not matter how much you have

<p>ln(A) = ln(A)<sub>0</sub> -kt</p><p>t1/2= 0.693/k — does not matter how much you have </p>
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2nd order rxn

1/(A) = 1/(A)0 + kt

t1/2 = 1/k(A)0 — the less of a reactant the longer it will take

<p>1/(A) = 1/(A)<sub>0</sub> + kt</p><p>t1/2 = 1/k(A)<sub>0</sub> — the less of a reactant the longer it will take </p>
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K >> 1

products favored at equilbrium

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K << 1

reactants favored at equilibrium

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Q < K

shift right

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Q > K

shift left

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Q = K

at equilibrium