Chem 1312, Exam 1

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30 Terms

1

Molarity

mol solute / L solution

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2

Molality

(moles of solute) / (kg of solvent)

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3

Mole Fraction

(# moles solute) / (Total # of moles)

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4

Mass Percent

(Mass of solute) / (Mass of solution) x100

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5

Henry’s Law

C = kP

where C: concentration of the dissolved gas

P: partial pressure of the gaseous solute

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6

Raoult’s Law (vapor pressure)

Psolution = Xsolvent . P*vapor pressure of pure solvent

if they ask for 2 just add the results together

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7

Boiling Point Elevation

changeT = Kbmsolute

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8

Freezing Point Depression

changeT = Kfmsolute

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9

Osmotic Pressure

piosmotic p = MRT

R: 0.08206

T in kelvin

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10

i

(moles of particles in solution) / (moles of solute dissolved)

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11

rate

(over the same temperature)

[A]t2 - [A]t1 / t2 - t1 = k[A]n

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12

Order of reaction

n + m + l ….

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13

Arrehenius Equation for 1

knowt flashcard image
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14

Arrehenius Equation for 2

knowt flashcard image
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15

0 order Rate Law

Rate = k

[A] = -kt + [A]0

T1/2 = [A]0 / 2k

units: M/S

<p>Rate = k</p><p>[A] = -kt + [A]<sub>0</sub></p><p>T<sub>1/2 </sub>= [A]<sub>0</sub> / 2k</p><p>units: M/S</p>
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16

1st Order Rate Law

rate = k[A]1

ln[A] = -kT + ln[A]0

T1/2 = 0.693 / k

units: s-1

<p>rate = k[A]<sup>1</sup></p><p>ln[A] = -kT + ln[A]<sub>0</sub></p><p>T<sub>1/2</sub> = 0.693 / k</p><p>units: s<sup>-1</sup></p>
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17

2nd Order rate law

rate = k[A]2

1/[A] = kt + 1/[A]0

T1/2 = 1/k[A]0

units: M-1S-1

<p>rate = k[A]<sup>2</sup></p><p>1/[A] = kt + 1/[A]<sub>0</sub></p><p>T<sub>1/2 </sub>= 1/k[A]<sub>0</sub></p><p>units: M<sup>-1</sup>S<sup>-1</sup></p>
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18

K

Equilibium Constant

on a reaction

jA + kB >< lC +mD

the law mass action = picture

<p>Equilibium Constant</p><p>on a reaction</p><p>jA + kB &gt;&lt; lC +mD</p><p>the law mass action = picture</p>
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19

reverse Kc

1/Kc

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20

Kc when there is a coefficient (n) multiplying the equation

Kcn

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21

pressure and concentraiton equation

P = CRT

C: concentration (M)

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22

Equilibrium of Partial Pressure in a gas

Kp

(same as Kc but with partial pressures)

<p>Kp</p><p>(same as Kc but with partial pressures)</p>
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23
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24

Kc in terms of Kp

K = Kp(RT)-changen

change in n = product coefficients - reactant coefficients

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25

Q = K

Equilibrium

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26

Q > K

shift to the left

consuming products and creating reactants

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27

Q < K

shift to the right

consuming reactants and creating products

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28

I.C.E shift to the right

-x of the reactants

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29

I.C.E shift to the left

-x to the products

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30

Quadratic Formula

x = (-b ± √(b²-4ac)) / (2a)

<p><strong><mark>x = (-b ± √(b²-4ac)) / (2a)</mark></strong></p>
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