CHEM 2080 Prelim 1

Final - initial

• Heat is not energy, it’s the transfer of energy. Not a state function. Thus, lowercase.
• q is positive if thermal energy is added to the system. (endothermic)

• Energy that is transferred to the system by force.
• w = force*distance
• w is positive if energy is transferred to the system by force. (compression)
• W = -force * Δh (displacement)
• Force = pressure X area
• Or w = -pressure (vfinal - vinitial) for cylinders.

ΔU = q + w

• Force/area
• Thus, force = pressure X area

• Pressure = pascal
• Work = joule
• Volume = m3
• An alternative is to use L.atm

Exothermic

Endothermic

• Functions that depend only on the state of the function and not on how the state was achieved. They can be calculated by final - initial
• Written in upper case. For example ΔU is a state function.
• So are P T and V.

It’s the same.

It’s equal to q.

It’s equal to q + w

• It’s the heat released in constant pressure reactions?
• ΔH = ΔU + P*Δv (remember that -P*ΔV is equal to w)
• Another form is ΔU + Δmolofgas *R*T

They’re the same unless gas is produced or consumed.

ΔU > ΔH

ΔH > ΔU

• Heat needed to raise the temperature of a 1 mol of a substance by 1K.
• j/molK
• Q = n * Cp * ΔT

• Heat needed to raise the temperature of 1 gr of a substance by 1K
• J / gK
• Q = m * Csp * ΔT

• The total Q will be 0 due to insulation.
• Keep in mind that the heat given and the heat received will be the same.
• Don’t forget units. Always write units. However, keep in mind that you should write K and C accordingly.

-ΔH of the reaction will be equal to the heat received by the water due to insulation.

Are exothermic.

Endothermic.

Run the experiment under constant volume conditions.

• Energy, enthalpy, entropy, free energy.
• Work and heat are NOT state functions.

ΔH is additive. Be careful with this it’s easy to mess up.

• These are tabulated.
• Enthalpy required to form one mole of a substance from the most stable form of the constituent elements at STP.
• Remember to convert from mol to gram etc.

Graphite.

0.

Sum of ΔH of formation of products - sum of ΔH of formation of reactants

ΔH = -cp(cal)*ΔT

• 1/stoichiometric coefficient [Dconc.]/Dt=1/…. Of all reactants and products.
• For reactants, it’s negative.

Draw the tangent line to the specific point. The absolute value of the slope is the rate.

The average rate is given by taking the initial value and the final value and dividing the difference by time.

• Equation that describes reaction rate in terms of reactant concentration.
• =k [A] [B] [C]
• ONLY REACTANTS

• Only experimentally.
• The key point here is that we must isolate the reaction of interest from the backward reaction. Thus, we use the initial rate!! Time is equal to 1.
• We must do 3 tests:
• Plot [A] to time
• Plot ln[A] to time
• Plot 1/[A] to time
• Whichever is linear will give you the reaction order.

Number of reactants + 1.

Equal to the sum of the exponents in the rate equation.

• [A] = [A]_0 - kt
• Slope = -k
• X intersect is [A]0

• [A] = [A]_0 * e^(-kt)
• Natural log gives you ln[A] = -kt + ln[A]0 (linear form!)
• When you plot ln[A] to time, it’ll give you a line going down.
• X intersect is ln[A]0
• Slope = -k

• 1/[A] = kt + 1/[A]0 (linear form)
• Plotting 1/[A] will give you a line going up.
• X intersect is 1/[A]0
• Slope = k

K

Number of experiments = number of unknowns.

First order reactions

Solve integrated rate law for [A]½

• Run the experiment 3 times with the 2 reactants at high concentration and monitor the reactant under study.
• Graph [A], ln[A], 1/[A] determine which is most linear.
• Analyze any of the data to determine k.

Target conc = [A]0 * (½)^(n/half life)

3-body collisions are extremely rare. The direct mechanism rarely occurs.

Geometric shape when everything is together.

When the temporary product can be isolated. Produced in one step of the reaction and consumed in another. Usually an unstable and reactive species.

Determine whether the intermediate is present.

Is determined by how many molecules have to clash into each other.

• The rate determining step (slowest one) can be used, as we’re certain that it occurs in one step.
• Take the stoichiometry.

• Reactants must collide:
• With enough energy
• With correct orientation

Rate = (rate of reactants colliding, which is proportional to concentration) x (prob, that molecules will have enough energy, depends on temperature) x (probability of correct orientation, which is a constant)

Ke must be bigger than Ea to occur. It will maybe happen.

K = A e ^(-EA/RT)

ln(k2/k1) = Ea/R x (T2-T1)/(T2*T1)

Provides a different and faster reaction mechanism.

If the presence of a molecule increases the rate of reaction and it isn’t consumed by the end, it’s a catalyst.

• No. It decreases the Ea of both the forward and the backward reactions.
• It’ll only bring the equation to equilibrium faster.

It’s the elementary reaction that occurs slower.