k = constant of proportionality called the rate constant

n = reaction order. The value of this determines how the rate depends on the concentration of the reactant

-zero order

- independent of the concentration of A

-first order

-directly proportional to the concentration of A

-second order

-proportional to the square of the concentration of A

if n = 2, the reaction is __, and the rate is __

-decreases

-constant

-does not

-slows down

- decreases

the initial rate is measured by running the reaction several times with different initial reactant concentrations to determine the effect of the concentration on the rate

-initial rate

- initial concentration

-same

-initial

-quadruples

-doubling

-quadratic

 For a __ reaction you can substitute any two initial concentrations and the corresponding initial rates into a ratio of the rate laws to determine the order (n): rate 2 / rate 1 = k[A]ⁿ₂ / k[A]ⁿ₁

 The rate constant for a zero-order reaction has units of __

The rate constant for a second-order reaction has units of __

-A and B

-m

-n

in the reaction order for multiple reactions,Rate = k[A]^m[B]ⁿ; 

__ are reactants

__ is the reaction order with respect to A

__ is the reaction order with respect to B

-experiment

-method of initial rates

[A]_t

k

[A]₀

for all of the integrated rate laws:

__ = the concentration of A at any time, t

__ = rate constant

__ = initial concentration of A

straight line

ln[A]_t = -kt + ln[A]_{0 -->}

y       = mx +  b

straight line

–k

ln[A]₀

negative

positive

straight line

 1 / [A]_t = kt + 1 / [A]_{0 -->}

y         = mx +  b

inverse

straight line

k

1 / [A]₀

straight line

[A]_t = -kt + [A]_{0 -->}

y     = mx +  b

straight line

–k

[A]₀

rate constant

initial concentration

reaction order

rate law

reactant(s)

time

The integrated rate law (which is mathematically derived from the rate law) relates the concentration of the __ to __

zero-order

second-order

A decomposition reaction, with a rate that is observed to slow down as the reaction proceeds, is found to have a half-life that depends on the initial concentration of the reactant. Which of the following is most likely to be true of this reaction?

A) a plot of the natural log of the concentration of the reactant as a function of time will be linear

B) the half-life of the reaction increases as the initial concentration increases

C) A doubling of the initial concentration of the reactants results in a quadrupling of the rate

increase

faster

A

-E_a

e^{-Ea / RT}

R

in the Arrhenius equation,

__ = constant called the frequency factor (or pre-exponential factor)

__ = activation energy (or activation barrier)

__ = exponential factor

__ = gas constant

successful

product

temperature (T)

activation energy (E_a)

one (e^-0 = 1)

successful

negative number

very small

Commonly, only a __ fraction of molecules have enough energy to make it over the activation barrier

increases

decreases

k = Ae^{-Ea / RT} = pze^{-Ea / RT}

p = orientation factor

z = collision frequency

if two molecules are to react with each other, they must collide in such a way that allows the necessary bonds to __

nuclei

weaken

form

products

Which reaction would you expect to have the smallest orientation factor?

A) H(g) + I(g) --> HI(g)

B) H₂(g) + I₂(g) --> 2 HI (g)

C) HCl(g) + HCl(g) --> H₂(g) + Cl₂(g)

unimolecular

bimolecular

1

k[A]

2

k[A]²

2

k[A][B]

3

k[A]³

3

k[A]²[B]

3

k[A][B][C]

limits

rate law

1. The elementary steps in the mechanism must sum to the overall reaction

2. The rate law predicted by the mechanism must be consistent with the experimentally observed rate law

concentration of the reactants

temperature

Use of solid catalysts with gas-phase or solution-phase reactants is the most common type of __

lowered

faster

which

when

specific

efficient

right

high

low

left

high

low

maintain equilibrium

counteract

larger

product

increase

decrease

occurring

occurring at the same rate

most

small fraction

Some reactions reach equilibrium only after __ of the reactants have formed products

Some reactions reach equilibrium when only a __ of the reactants have formed products

slows down

speeds up

not equal

equal

quantify

concentrations

ratio

products

reactants

molarity of products

molarity of reactants

molar concentration of A (M; molarity)

in the equilibrium constant (K) equation (also known as the law of mass action) -->K = [C]^c[D]^d / [A]^a[B]^b

[C]^c[D]^d = __

[A]^a[B]^b = __

[A] = __

balanced chemical equation

law of mass action

far to the right

products

reactants

far to the left

reactants

products

(b) the reaction mixture will contain 1 mol of A and 10 mol of B so that [B]/[A] = 10

...it will contain more moles of B when compared to A because the equilibrium constant (K) is 'large' so the forward reaction is favored and the concentration of products is large and the concentration of reactants is small

The equilibrium constant for the reaction A (g) <--> B (g) is 10. A reaction mixture initially contains 11 mol of A and 0 mol of B in a fixed volume of 1 L. When equilibrium is reached, which statement is true?

(a) the reaction mixture will contain 10 mol of A and 1 mol of B

(b) the reaction mixture will contain 1 mol of A and 10 mol of B

(c) the reaction mixture will contain equal amounts of A and B

1. reversing the equation

2. multiplying the coefficients in the equation by a factor

3. adding two or more individual chemical equations to obtain an overall equation

0

equal

reactants

products

For which reaction does K_p = K_c

(a) 2Na₂O₂(s) + 2CO₂(g) « 2Na₂CO₃(s) + O₂(g)

(b) NiO(s) + CO(g) « Ni(s) + CO₂(g)

(c) NH₄NO₃(s) « N₂O(g) + 2H₂O(g)

moles per liter (M)

unitless

equilibrium concentrations

ICE table

equilibrium constant

reactant quotient

The difference between the reaction quotient and the equilibrium constant is that, at a given temperature, the __ has only one value and it specifies the relative amounts of reactants and products at equilibrium

The __ depends on the current state of the reaction and has many different values as the reaction proceeds

less than

goes to the right (toward products)

goes to the left (toward reactants)

is at equilibrium

Q < K ; reaction __

Q > K ; reaction __

Q = K ; reaction __

  1. Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products (ICE table)

2. Use the initial concentrations to calculate the reaction quotient (Q) for the initial concentrations. Compare Q to K to predict the direction in which the reaction will proceed.

3. Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products in terms of x

4. Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x

5. Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Using the given value of the equilibrium constant, solve the expression for the variable x.

6. Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations

7. Check your answer by substituting the computed equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K

 1. Using the balanced equation as a guide, prepare a table showing the known initial partial pressures of the reactants and products

2. Use the initial partial pressures to calculate the reaction quotient (Q). compare Q to K to predict the direction in which the reaction will proceed

3. Represent the change in the partial pressure of one of the reactants or products with the variable x. Define the changes in the partial pressures of the other reactants or products in terms of x

4. Sum each column for each reactant and product to determine the equilibrium partial pressures in terms of the initial partial pressures and the variable x

5. Substitute the expressions for the partial pressures (from step 4) into the expression for the equilibrium constant. Use the given value of the equilibrium constant to solve the expression for the variable x

6. Substitute x into the expressions for the equilibrium partial pressures of the reactants and products (from step 4) and calculate the partial pressures

7. Check your answer by substituting the calculated equilibrium partial pressures into the equilibrium expression. The calculate value of K should match the given value of K

1. Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products

2. Use the initial concentrations to calculate the reaction quotient (Q). Compare Q to K to predict the direction that the reaction will proceed

3. Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products with respect to x

4. Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x

5. Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Use the given value of the equilibrium constant to solve the expression for the variable x.

in this case, the resulting equation is cubic in x. Although cubic equations can be solved, the solutions are not usually simple. However, since the equilibrium constant is small, we know that the reaction does not proceed very far to the right. Therefore, x will be a small number and can be dropped from any quantities in which it is added to or subtracted from another number (as long as the number itself is not too small)

check whether your approximation was valid by comparing the calculated value of x to the number it was added to or subtracted from. The ratio of the two numbers should be less than 0.05 ( or 5%) for the approximation to be valid. If approximation is not valid proceed to step 5a

5a. if the approximation is not valid, you can either solve the equation exactly (by hand or with your calculator), or use the method of successive approximations. In this case, we use the method of successive approximations

 substitute the value obtained for x in step 5 back into the original cubic equation, but only at the exact spot where x was assumed to be negligible and then solve the equation for x again. Continue this procedure until the value of x obtained from solving the equation is the same as the one that is substituted into the equation

6. Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations

7. Check your answer by substituting the calculated equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K. Note that the approximation method and rounding errors could cause a difference of up to about 5% when comparing values of the equilibrium constant

For the generic reaction, A(g) « B(g), consider each value of K and initial concentration of A. For which set will the x is small approximation most likely apply?

(a) K = 1.0 x 10^-5; [A] = 0.250M

(b) K = 1.0 x 10^-2; [A] = 0.250M

(c) K = 1.0 x 10^-5; [A] = 0.00250M

(d) K = 1.0 x 10^-2; [A] = 0.00250M

<

right (in the direction of the products)

>

left (in the direction of the reactants)

>

left (in the direction of the reactants)

<

right (in the direction of the products)

inversely

increase

decrease

increases

do not change

no

does not shift in either direction