A)   The order of the reaction with respect to substance B is 1

A)   Substance B is not involved in any of the steps in the mechanism of the reaction

A)   Substance B is not involved in the rate-determined step of the mechanism, but is involved in subsequent steps

A)   Substance B is probably a catalyst, and as such, its effect on the rate of the reaction does not depend on its concentration

Ce⁴⁺ and Tl+

Ce³⁺ and Tl³⁺         

Ce³⁺ and Mn³⁺           

Ce³⁺ and Mn⁴⁺          

Tl³⁺ and Mn²⁺  

Rate = k [NO₂] [CO]        

Rate = k [NO₂]²      

Rate = k [NO₃] [CO]       

Rate = k [NO₂] [NO₃] [CO]

Rate = k[Br₂] [NO]         

Rate = k[Br₂] [NO]²       

Rate = k[Br₂]² [NO]       

Rate = k[NOBr₂] [NO]

The spark provides the activation energy required to get the reaction started

The hot chemicals from the flint catalyze the reaction

The heat from the spark increases the activation energy for the reaction

The heat from the spark decreases DE, making the reaction less exothermic

The heat from the spark removed energy from the bonds in the methane molecules, thereby lowering the activation energy for the forward reaction.

I only

II only

I and II only

I and III only

I, II, and III

A catalyst is present

The reaction order can be obtained directly from the balanced equation

The reaction order is zero

$$\Delta$$ S for the reaction is zero

$$\Delta$$ H for the reaction is zero

Molecules must collide in the correct orientation with enough energy to bond.

Molecules need enough energy to collide and react.

Atoms constantly collide and react.

The minimum energy needed for atoms to react

How much energy is needed for a reaction to occur.

The energy required to break a bond.

The time it takes for a reaction to occur.

Collision Theory

Not affect the rate of reaction.

Speed up the rate of reaction

Slow down the rate of reaction

Large chunks of chalk

Cube of sugar

Powdered sugar

Small chunks of sugar

The reaction rate will stay the same

The reaction rate will speed up

The reaction rate will slow down

Molecules react by colliding together

The effective collisions must occur with certain minimum amounts of energy

In a large sample, the greater the number of effective collisions, and the faster the rate of reaction

All of these are correct

Decrease the concentration of the acid

crush the chalk to increase the surface area

Put the test tube in an ice bath

Decreases, because there are LESS possible sites for correct collisions

Increases, because there are LESS possible sites for correct collisions

Decreases, because there are MORE possible sites for correct collisions

Increases, because there are MORE possible sites for correct collisions

to slow down

move faster

collide higher

in the right order

By lowering the activation energy or reaction

by giving them more energy

by making them more available

A

B

A

B

Slow down the reaction

Speed up the reaction

Have no affect on the reaction

unimolecular

bimolecular

polimolecular

trimolecular

unimolecular

bimolecular

trimolecular

polimolecular

rate = k [A]² [B]

rate = k [A]² [B]²

rate = k [A] [B]

rate = k [A] [B]²

rate = k [N₂O]²

rate = k [N₂O]

rate = k [N₂O]² [N₂]

rate = k

The products have more energy then the reactants.

The reactants have more energy then the products.

SO₃(g) + H₂(g) --> SO₂(g) + H₂O(g)

N₂(g) + 3H₂(g) --> 2NH₃(g)

Ag⁺(aq) + Cl^-(aq) --> AgCl(s)

C₂H₂(g) + 2H₂(g) --> C₂H₆(g)

MgSO₃(s) --> MgO(s) + SO₂(g)

$$\Delta$$ H° > 0 and  $$\Delta$$  S° > 0     

$$\Delta$$ H° > 0 and  $$\Delta$$  S° < 0      

$$\Delta$$  H° < 0 and  $$\Delta$$  S° > 0    

$$\Delta$$  H° < 0 and  $$\Delta$$  S° < 0

C₂H₅OH(l)  -->  C₂H₅OH(g)  

NaCl(s)  -->  NaCl(l)

CO₂(s)  -->  CO₂(g)

Cl₂(g)  -->  Cl₂(l)

$$\Delta$$  H = positive

$$\Delta$$  S = positive

$$\Delta$$  H = positive

$$\Delta$$  S = negative

$$\Delta$$  H = positive

$$\Delta$$  S = equal to zero

$$\Delta$$  H = negative

$$\Delta$$  S = positive

$$\Delta$$  H = negative

$$\Delta$$  S = negative

It must be positive, since the reaction is thermodynamically unfavorable at 600 K.

It must be negative, since there are more moles of products than reactants

It must be positive, since $$\Delta$$  G° is negative and $$\Delta$$  H° is positive

It must be negative, since $$\Delta$$  G° is positive and $$\Delta$$  H° is positive

$$\Delta$$ H° > 0 and $$\Delta$$ S° > 0       

$$\Delta$$ H° > 0 and $$\Delta$$ S° < 0     

$$\Delta$$ H° < 0 and $$\Delta$$ S° > 0     

$$\Delta$$ H° < 0 and $$\Delta$$ S° < 0    

+170 J/(mol_rxn×K)         

-170 J/(mol_rxn×K) 

+190 J/(mol_rxn×K) 

-190 J/(mol_rxn×K) 

equal to zero

equal to $$\Delta$$ H°         

greater than $$\Delta$$ H°

less than $$\Delta$$ H°

They are completely independent of each other; temperature has no effect on entropy

There is a direct relationship because at higher temperatures there is an increase in energy dispersal

There is an inverse relationship because at higher temperatures substances are more likely to be in a gas state

It depends on the specific gas and the strength of the intermolecular forces between individual molecules

H

G

S

E

accuracy

precision

the disorder of a system

the attraction of a nucleus for an electron

sublimation of CO₂

boiling water

condensation of steam on cold glass

NaCl dissolving in water

CaCO₃(s) → CaO(s) + CO₂(g)

3O₂(g) → 2O₃(g)

2NH₃(g) → 3H₂(g) + N₂(g)

C₆H₆(l) → C₆H₆(g)

a molecule is broken into two or more smaller molecules

a reaction occurs that results in an increase in the number of moles of gas

a solid changes to a liquid

all of these

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

H₂O(g) + CO₂(g) → H₂CO₃(aq)

H₂(g) + I₂(g) → 2Hl(g)

C₂H₂O₂(g) → 2CO(g) + H₂(g)

lower energy and less disorder

lower energy and more disorder

higher energy and less disorder

higher energy and more disorder

1 mole of KNO₃(l)

1 mole of KNO₃(s)

1 mole of H₂O(l)

1 mole of H₂O(g)

Molecular disorder increases

Molecules increase in number from solid to gas

Molecules are heavier in solid

Molecules are more reactive

solid

liquid

gas

aqueous

True

False

CuCO₃ (s) + 2HCl (aq) → CuCl₂ (aq) + CO₂ (g) + H₂O (l)

C (s) + O₂ (g) → CO₂ (g)

CuO (s) + H₂SO₄ (aq) → CuSO₄ (aq) + H₂O (l)

BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaCl (aq)

slow process

fast process

process that needs an external intervention to occur

process that does not need external intervention to occur / keep happening

Evaporating Water

Cooling down of hot water

Condensing Water Vapor

Solidifying of Magma

Evaporating Water

Cooling down of hot water

Condensing Water Vapor

Solidifying of Magma

Greater than zero

Lesser than zero

Equal to zero

Entropy wiill not be equal to zero, so its either of the two.

ΔS = +ve and non spontaneous

more order

more randomness

increase

decrease

ice melting

salt dissolving

water heating up

none of these

Building a skyscraper

A clean room becomes cluttered

An igloo melting

Osmosis (I hope you know your Biology!)

+ΔH and +ΔS

+ΔH and -ΔS

-ΔH and -ΔS

-ΔH and +ΔS

+ΔH and +ΔS

+ΔH and -ΔS

-ΔH and -ΔS

-ΔH and +ΔS

SO_3(g)

SO_2(g)

Br_2(l)

Br_2(g)

Sn_(s)

Pb_(s)

Entropy is increasing in the forward direction

Entropy is decreasing in the forward direction

Entropy increases in the forward direction

Entropy decreases in the forward direction

Equilibrium is achieved with a reasonable amount of reactants and products

Equilibrium goes almost to completion

Equilibrium virtually does not occur

will achieve equilibrium with a reasonable amount of reactants and products

Will go almost to completion

Will virtually not ocur

reaction achieves equilibrium with a reasonable amount of reactants and products

reaction virtually does not occur

reaction goes virtually to completion

The N₂(g) and the H₂(g) must be mixed in a 1:3 ratio for a reaction to occur

A high activation energy makes the forward reaction extremely slow at 298 K

The reaction has an extremely small equilibrium constant, thus almost no product will form

The reverse reaction has a lower activation energy than the forward reaction, so the forward reaction does not occur

$$\Delta$$ G° > 0 because energy is released as the bond between H and Cl atoms forms, and entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles.

$$\Delta$$  G° > 0 because energy is absorbed as the bond between H and Cl atoms forms, and entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles.

$$\Delta$$  G° < 0 because energy is absorbed as the bond between H and Cl atoms forms, and entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles.

$$\Delta$$  G° < 0 because although entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles, energy is released as the bond between the H and Cl atoms forms.

Is only thermodynamically favorable in the presence of a catalyst

Occurs with an increase in entropy because O₂(g) is a product

Is reversible and H₂O₂(aq) is produced almost as fast as it decomposes

Has a slow rate at 298 K because the activation energy is relatively high

Because $$\Delta$$ G° > 0, K_a > 1, and HA completely dissociates

Because $$\Delta$$ G° > 0, K_a > 1, and HA almost completely dissociates

Because $$\Delta$$ G° > 0, K_a < 1, and HA only partially dissociates

Because $$\Delta$$ G° > 0, K_a < 1, and HA does not dissociate

$$\Delta$$ G° = -RTlnK > 0 because K > 1         

$$\Delta$$ G° = -RTlnK < 0 because K > 1  

$$\Delta$$ G° = $$\Delta$$ H° - T $$\Delta$$ S° < 0 because $$\Delta$$ H° < 0 and $$\Delta$$ S° > 0

$$\Delta$$ G° = $$\Delta$$ H° - T $$\Delta$$ S° > 0 because $$\Delta$$ H° < 0 and $$\Delta$$ S° < 0

All endothermic processes are thermodynamically favorable

Stirring the solution during dissolution adds the energy needed to drive an endothermic process

Dissolving the salt decreases the enthalpy of the system

Dissolving the salt increases the enthalpy of the system

The particle diagrams best represent that $$\Delta$$ H° < 0 because the ions from both compounds are solvated by water molecules

The particle diagrams best represent that $$\Delta$$ H° < 0 because both compounds produce about the same amount of CO₃^2-ions from the dissolution

The particle diagrams best represent that $$\Delta$$ S° > 0 because both compounds produce a very small amount of ions from the dissolution

The particle diagram best represent that the molar solubility is greater for FeCO₃ compared to MnCO₃

I only

I and II only

I and III only

II and III only

I, II, and III

How much energy is given off by a reaction.

The tendency of a reaction to become "random"

How spontaneous a reaction is.

Enthalpy of the reaction.

Entropy of the reaction.

The temperature of the reaction in Kelvin.

All of the above are needed.

A high negative enthalpy.

A low negative enthalpy.

A high positive enthalpy.

A low positive enthalpy.

0

4

8

low

high

very high

very low

+ΔH and +ΔS

+ΔH and -ΔS

-ΔH and -ΔS

-ΔH and +ΔS

1

3

5

-85.6 kJ, spontaneous

-18.3 kJ, not spontaneous

+18.3 kJ, spontaneous

+85.6 kJ, not spontaneous

ΔG + ΔH - ΔS -

ΔG - ΔH - ΔS -

ΔG - ΔH + ΔS +

ΔG - ΔH + ΔS +

Δ H = - , Δ S = + and Δ G = - (all temperature)

Δ H = + , Δ S = + and Δ G = - (At all temperature)

Δ H = + , Δ S = + and Δ G = - (At low temperature)

Δ H = + , Δ S = + and Δ G = - (At high temperature)

452 K

2210 K

382 K

363 K

-252.6 kJ

-13.4 kJ

13.4 kJ

252.6 kJ

A

B

C

D

A

B

C

D

A

B

C

D

A

B

C

D

E

positive, positive

positive, negative

negative, negative

negative, positive

positive, positive

positive, negative

negative, negative

negative, positive

positive, positive

positive, negative

negative, negative

negative, positive

positive, positive

positive, negative

negative, negative

negative, positive

Zero

Positive

Negative

zero

positive

negative

calorie

kelvin

joule

watt

reaction is exothermic

reaction is endothermic

reaction is spontaneous

reaction is not spontaneous

enthalpy increases

enthalpy decreases

entropy decreases

entropy increases

reaction is exothermic

reaction is endothermic

reaction is spontaneous

reaction is not spontaneous

A spontaneous reaction

A nonspontaneous reaction

True

False

a nonspontaneous reaction.

a spontaneous reaction.

True

False

True

False

True

False

The catalyst could change the reaction from second order to third order

The catalyst could increase the particles’ speed, thereby increasing the value of the rate constant, k

The catalyst could decrease the particles’ speed, making it less likely that the particles will rebound without reacting when they collide

The catalyst could absorb one of the particles, making a successful (reaction-producing) collision with the other particle more likely

I only

II only

III only

I and II only

I, II, and III

A

B

C

D

none of these

The diagram shows no indication of any activation energy

The forward and reverse activation energies are equal

The forward activation energy would be greater

The reverse activation energy would be greater

none of these

The products would be produced at a lower energy state

The rate of the reaction would tend to increase

The activated complex would convert into products

The reactants would continue to exist in their present state

The products would form at an unstable energy state

2O₃(g)  +  2NO(g) -->  4O₂(g)  +  N₂(g) (slow) 

O₃(g) + NO(g) -->  NO₂(g) + O₂(g) (slow)

NO₂(g) + O(g)  -->  NO(g) + O₂(g) (fast)

NO₂(g) + O₃(g)  -->  NO(g) + 2O₂(g) (slow)

NO(g)  +  O(g)  -->   NO₂(g)  (fast)

NO₂(g) + O(g)  -->  NO₃(g)  (slow)

NO₃(g) + O₃(g)  -->  NO₂(g) + 2O₂(g) (fast)

It promotes the proper orientation of N and H atoms to form new N-H bonds

It provides a reaction path with a lower activation energy

It forms a bond between Ru and N that is stronger than the bond in N₂.

It decreases the frequency of collisions between N₂(g) and H₂(g)

Curve C will broaden and Line B will move to the left, because more Y₂ molecules will have an energy greater than the minimum energy needed to overcome the activation energy barrier

Line B will move to the left because a larger fraction of the Y₂ molecules will have the minimum energy to overcome the activation energy barrier

Curve C and Line A will move to the right because the average energy of the Y₂ molecules will increase

Line A will move to the right because the most probable energy of the Y₂ molecules will increase

the relative difference in temperature of reactants and products

the relative difference in amounts of reactants and products

the relative difference in energy of reactants and products

the relative difference in volume of reactants and products

acting energy

active energy

activation energy

Endothermic

Exothermic

Endothermic

Exothermic

A

B

C

The energy needed to start a reaction

The temperature needed to start a reaction

The minimum energy that particles must have to react

The maximum energy that particles must have to react.

Exothermic

Endothermic

By lowering the activation energy or reaction

by giving them more energy

by making them more available

A

B

C

D

E

a

b

c

A

B

C

ALL THE ABOVE

1

2

True

False

Platinum

Phosphorous

Palladium

Phosphorous and platinum

Platinum and palladium

Reaction profile

Energy graph

Catalysing reaction graph

Overall change of energy grapg

Bio-chemical reactions are mostly catalysed by enzymes

Catalyst does not start a reaction

Catalyst changes the equilibrium constant of a reaction

Co-enzymes increase the activity of an enzyme

in the process of adsorption, the activation energy of the molecules becomes large.

adsorption produces heat which increases the speed of the reaction

the concentration of the reactant molecules at the active centres of the catalyst becomes high due to adsorption.

adsorption lowers the activation energy of the reaction

Once - it gets used up in the reaction

A few times

None

Lots of times - it does not get used up in the reaction

True

False

Curve II is at the lower temperature and there are less molecules in the sample

Curve II is at the lower temperature and there are the same number of molecules in the sample

Curve II is at the higher temperature and there are more molecules in the sample

Curve II is at the higher temperature and there are the same number of molecules in the sample

Black (peak on the left)

Red (peak in the middle)

Blue (peak on the right)

All particles are moving at the same speed

The area under the curve to the left of Ea decreases.

The total area under the curve increases.

The activation energy decreases.

More molecules have the mean energy.

Position A represents the mean energy of a molecule in the container.

Addition of a catalyst moves the position of E_A to the right.

The area under the curve to the right of E_A represents the number of molecules with enough energy to react.

The position of the peak of the curve at a higher temperature is further away from both axes.

1 only

1 and 2

2 and 3

3 only

W, X and Z only

W, Y and Z only

W and Z only

X and Y only

decreasing the size of the solid copper particles

increasing the temperature of HNO₃ by 20 °C

increasing the concentration of HNO₃

allowing NO₂ gas to escape

Reaction 1 < Reaction 2 < Reaction 3 < Reaction 4.

Reaction 3 < Reaction 1 < Reaction 4 < Reaction 2.

Reaction 4 < Reaction 2 < Reaction 1 < Reaction 3.

Reaction 3 < Reaction 1 < Reaction 2 < Reaction 4.

the whole graph will shift to the right meaning that more molecules will have enough energy to react

the activation energy shifts to the right meaning that more molecules will have enough energy to react

the activation energy shifts to the left meaning that more molecules will have enough energy to react

the activation energy line stays where it is but the height of the graph at the activation energy point will increase meaning that more molecules will have enough energy to react

particles that can react at the lower temperature

particles that can react at the higher temperature

particles that cannot react at the lower temperature

particles that cannot react at the higher temperature

the white area is smaller than the dark blue area

the white area is larger than than the dark blue area

the white area is equal to the dark blue area

T1 is hotter than T2

beautiful neon green stripey baby

quantity of particles with a kinetic energy greater than the activation energy prior to the addition of the catalyst

quantity of particles with a kinetic energy greater than the activation energy after to the addition of the catalyst

quantity of particles with a kinetic energy greater than the activation energy prior to the temperature change

Z - X

Y

X

Z-Y

I and II only

I and III only

II and III only

I, II and III

ΔH is z, E_a (catalysed reaction) is x + z

ΔH is z, E_a (catalysed reaction) is z + y

ΔH is -z, E_a (catalysed reaction) is x

ΔH is z + x, E_a (catalysed reaction) is x