# Exam 1 Key Equations and Problems

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- Wisconsin
- University of Wisconsin - Eau Claire
- Chemistry
- Chemistry 104
- King
- Exam 1 Key Equations and Problems

**Created:**2012-02-20

**Last Modified:**2012-02-20

^{2 }

**OR**E

_{KE}=(1/2)mv

^{2}

**OR**w=Fd

**OR**E

_{PE}=mgh

c=q/m(Δ**t**) **OR **q=mc(changein**t**)

*note the lowercase c

where:

q=heat

m=mass

t=temp

Equation for heat capacity:

C=q/(Δ**t**)

*note the uppercase C

where

q=heat

t=temp

**calorimetry**?

1. heat changes measured under **constant pressure** conditions in a device **open to the atmosphere**

ΔH= q_{p}

2. heat changes measured under **constant volume** conditions in a **sealed container**

ΔE = q_{v}

*E=internal energy

H = E + p V

(enthalpy = internal energy + (pressure x volume))

(mol/liter)/second = mol l^{-1}s^{-1} = **Ms ^{-1}**

Rate of reaction = Δ[reactant] / time interval

**OR **rate = Δ[reactant] / Δt

* note that t in this case stands for time interval, NOT temperature

Include the stoichiometric coefficients of the reactant if the coefficients do not equal unity.

Rate of substance A = -(1/a)(Δ[A]/Δt)

*t= time interval

*note that we need a negative sign when finding the rate of a reactant

Rate of substance C = (1/c)(Δ[C]/Δt)

*t= time interval

*note that we do NOT need a negative sign when finding the rate of a product

rate = k[reactant]^{n}

*n=**determined by experiment**

*k=rate constant

If an equation is: aA + bB --> cC + dD, what is the **forward rate law**?

rate = k[A]^{x}[B]^{Y}

*X and Y are determined through experiment

*k = rate constant

^{1}[B]

^{2}, what order is the reaction in?

**first-order**in species A and

**second-order**in species B, therefore it is

**third-order over all.**

_{0}?

[A]_{t} = [A]_{0} - kt

*[A]_{t} = concentration of A at time t

*[A]_{0} = concentration of A at t=0 (initial concentration of A)

*k = rate constant

*t = time

_{t}= [A]

_{0}- kt , what variable in the rate would be equivilent to the slope(y)? (compare to y=mx + b) What would y, m, and x be?

slope(y) = -k

x = t <-- x-axis

y= [A]_{t }<-- y-axis

b= [A]_{0} <--how far you would move up on the y axis

**ln([A]**, then we use log properties to re-write again as:

_{0}/[A]_{t}) = kt**[A]**, then re-write as

_{t}/[A]_{0}= e^{-kt}**[A]**

_{t}= [A]_{0}e^{-kt}_{0}/[A]

_{t}) = kt , which variables would be equivalent to y, m, x, and b of the straight line equation?

y = ln([A]_{0}/[A]_{t})

m(slope) = k

x = t

^{2}, which can be re-written as

**1/[A]**

_{t}= 1/[A]_{0 }+ kt_{t}= 1/[A]

_{0}+ kt , which variables would be equivalent to y, m, x, and b of the straight line equation?

y = 1/[A]_{t}

m(slope) = k

x = t

b = 1/[A]_{0}

t_{1/2} = [A]_{0}/2k

_{1/2}= ln(2)/k

k = Ae^{-Ea/RT}

*E_{a} = Activation Energy

*k = rate constant

*R = gas constant

*T = temp (kelvin)

*A = frequency factor (related to collision frequency)

ln(e) = 1

ln(XY) = ln(X) + ln(Y)

ln(X/Y) = ln(X) - ln(Y)

ln(X^{m}) = m ln(X)

ln(e^{y}) = Y ln(e) = Y

ln*k* = ln(Ae^{-Ea/RT}) , which is then written as: **ln k = lnA - (E_{a}/RT) **

_{a}/RT) , which variables are equivalent to y, m, x, and b in the straigh line equation?

y = lnk

m = -(E_{a}/R)

b = lnA

x = T^{-1}

_{a}to E

_{a+c}(activation energy with catalyst), how would you set up an equation to compare the rate constants?

__k___{c} = __Ae ^{-Ea,c/RT}__

k = Ae ^{-Ea/RT}

^{}

Then re-write it all as **k _{c}/k = e^{(Ea-Ea,c)/(RT)}**

**overall**rate constant in the presense of a catalyst?

_{overall }= k + k

_{c}

reaction quotient = (concentration of products)^{product coeff.}/ (concentration of reactants)^{reactant coeff.}

**OR** Q = [products]^{product coeff.}/ [reactants]^{reactant coeff.}

*Q = reaction quotient

What is the reaction quotient for: aA + bB --> cC + dD ?

Q = [C]^{c}[D]^{d}/ [A]^{a}[B]^{b}

equilibrium constant (K)

*equation is the same as the reaction quotient (Q) equation, but Q is replaced with K

Better way to write K:

K = (([C]/M)^{c}([D]/M)^{d}) / (([A]/M)^{a}([B]/M)^{b})

*K is dimentionless!

_{c}= [C] / [A][B]

_{p}= p

^{2}

_{C}/ (p

_{A})(p

_{B})

_{c}to K

_{p}? Hint: Use the ideal gas law! (pv=nrt)

Plug each into the ideal gas law to make:

P_{A} = [A]RT and P_{B} = [B]RT

Then, plug the results into the expression for K_{p}

n_{A}/V_{B}=[A] and n_{B}/V_{B} = [B]

Then, re-write as K_{p} = [B]^{b}/[A]^{A} x (RT)^{Δn}

where Δn = b-a

therefore, **K _{p} = K_{C}(RT)^{Δn}**

_{3 }(s) <--> CaO (s) + CO

_{2}(g)

K_{c} = [CO2]

*Pure solids and liquids do not appear in the final equilibrium constant expression

K_{p} = p_{CO2}

A + B <--> E + F

K_{c} = [E][F]/[A][B]

ax^{2} + bx + c = 0

**OR**

x = (-b +- (SQ ROOT OF (b^{2} - 4ac)) / 2a

1. Concentration of molarities

2. Change Line

3. Equilibrium Line (molarity + change line)

* reactants on left, products on right

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