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PSU Chem 112 First Exam With Complete Solutions Latest Update
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PSU Chem 112 First Exam With Complete Solutions Latest Update
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Solution 2024/2025Pepper
PSU Chem 112 First Exam With Complete
Solutions Latest Update
formula for rate ANS✔✔ ∆concentration/∆time
effect of concentration on rate of reaction ANS✔✔ increased concentration means an increase in the probability of
having more collisions, increasing the rate of reaction
effect of temperature on rate of reaction ANS✔✔ molecules must collide with enough kinetic energy to react, increasing
the temperature increases the kinetic energy which makes collision more likely and increases the chance that a collision
will cause a reaction | increasing the temperature will increase the rate of reaction
rate of an a->b reaction ANS✔✔ -∆[A]/∆t = ∆[B]/∆t | reactants are positive
slope of a rate graph ANS✔✔ ∆[A]/∆t
rate of aA->bB ANS✔✔ (-1/a)(∆[A]/∆t) = (1/b)(∆[B]/∆t)
ratio of stoichiometric coefficient
rate law ANS✔✔ expresses the rate as a function of change in concentration over change of time
rate law for aA+bB->cC+dD ANS✔✔ rate=k[A]^m * [B]^n
reaction order ANS✔✔ show how the rate is affected by the reactant concentration
three reaction orders ANS✔✔ 0= change concentration and it doesn't affect rate
, Solution 2024/2025
Pepper
1= double concentration double rate
2= double concentration quadruple rate
overall order ANS✔✔ orders all summed up
units of k for zero first second and third order ANS✔✔ zero= ms⁻¹
first=s⁻¹
second=m⁻¹s⁻¹
third=m⁻²s⁻¹
find order in respect to a reactant equation ANS✔✔ (rate x / rate y)= ([ ]x / [ ]y)^m
find order of reactant ANS✔✔ find two experiments where only concentration is changing is the one youre looking for
then use the comparison equation
find the rate constant k ANS✔✔ find both orders of both reactants then pick an experiment plug in all the values and
solve for k
integrated rate laws ANS✔✔ used to find the time it takes to use up a specific amount of a reactant and the amount of
reactant used up in a specific time interval
first order integrated rate law ANS✔✔ ln[A]=ln[A]₀ -kt
second order integrated rate law ANS✔✔ 1/[A]t = 1/[A]₀ +kt
graph of what vs time is a first order ANS✔✔ ln[A]
graph of what vs time is a second order ANS✔✔ 1/[A]
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