Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

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

Abbey W.

Equation for Kinetic Energy:

Kinetic Energy = (1/2)mass x velocity^{2 }**OR **E_{KE}=(1/2)mv^{2}

Equation for Work:

Work = force x distance **OR **w=Fd

Advertisement

Equation for Potential Energy:

Potential Energy = mass x (accel. due to gravity) x height **OR **E_{PE}=mgh

Equation to find Change in Enthalpy:

Change in Enthalpy = Enthalpy of the products - Enthalpy of the reactants

Hess's Law: How to use it:

Reverse the first reaction and add it to the 2nd reaction. Cancel out. Write total reaction using remains. Add together enthalpies of the 2 reactions to find the enthalpy of the final reaction.

Equation for specific heat:

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

What are the 2 principal types of heat that can be measured in** 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

Formula used to connect enthalpy and interal energy formulas

H = E + p V

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

What is the most common unit used when measuring reaction rate?

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

Equation for the rate of reaction:

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.

Advertisement

If an equation is: aA + bB --> cC +dD, what is the rate equation for substance A?

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

If an equation is: aA + bB --> cC +dD, what is the rate equation for substance C?

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

Equation for the rate law:

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

If a rate law is: rate = k[A]^{1}[B]^{2} , what order is the reaction in?

The reaction is** first-order** in species A and **second-order** in species B, therefore it is **third-order over all.**

When a zero-order rate law is: -(Δ[A]/Δt) = k , how can we re-write this using [A]_{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

When looking at: [A]_{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

How would you write a first order rate law for A --> P?

-(Δ[A]/Δt) = k[A] , which can then be re-written as: Δ[A]/Δt = -k[A] , which can then be re-written as: **ln([A]**_{0}/[A]_{t}) = kt , then we use log properties to re-write again as: **[A]**_{t}/[A]_{0} = e^{-kt} , then re-write as **[A]**_{t} = [A]_{0}e^{-kt}

When looking at the equation: ln([A]_{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

If the reaction: A --> P , follows second-order kinetics, what is the rate law?

-Δ[A]/Δt = k[A]^{2} , which can be re-written as **1/[A]**_{t} = 1/[A]_{0 }+ kt

When looking at the equation: 1/[A]_{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}

What is the half-life equation for a zero-order reaction?

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

What is the half-life equation for a first-order reaction?

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

According to the collision theory of chemical kinetics, what can we expect rate to be directly proportional to?

(# of collisions) / sec

What is the Arrhenius equation?

k = Ae^{-Ea/RT}

*E_{a} = Activation Energy

*k = rate constant

*R = gas constant

*T = temp (kelvin)

*A = frequency factor (related to collision frequency)

Important Log properties:

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

How can the Arrhenius equation be re-written using natural logs (ln)?

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

When observing the Arrhenius equation re-written: lnk = lnA - (E_{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}

When comparing E_{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)}**

What is the reaction for the **overall** rate constant in the presense of a catalyst?

k_{overall }= k + k_{c}

What is the equation for a reaction quotient?

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}

When a system is at equilibrium, instead of measuring the reaction quotient (Q), we would measure the __________?

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!

When using: A + B --> C , how would you write the equation in terms of the equilibrium constant using concentrations of each species?

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

When using: A + B --> 2C , how would you write the equation in terms of the equilibrium constant using partial pressures of each speices?

K_{p} = p^{2}_{C} / (p_{A})(p_{B})

When using: a A <--> b B , how would you relate K_{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}**

T OR F: [Substance] = Density of the substance/ molar mass of the substance

TRUE

What will the final equilibrium constant expression for this equation be? CaCO_{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}

When adding these 2 reactions: A + B <--> C + D and C + D <--> E + F , what is the reaction and the corresponding equilibrium constant?

A + B <--> E + F

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

What is the quadratic equation?

ax^{2} + bx + c = 0

**OR**

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

What does an ICE table include?

1. Concentration of molarities

2. Change Line

3. Equilibrium Line (molarity + change line)

* reactants on left, products on right

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly! I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU
StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2016 StudyBlue Inc. All rights reserved.

© 2016 StudyBlue Inc. All rights reserved.