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Chapter 17 Question Bank

Total questions: 108

Worksheet time: 2hrs 39mins

Name
Class
Date
1.

Match the following keywords to their meaning:

a)

Explains the effects of temperature and volume on gas molecules; entropy increases when energy is more dispersed.

1.

Kinetic Molecular Theory (KMT)

b)

Occurs without external intervention; can be fast or slow

2.

Spontaneous Process

c)

In any spontaneous process, there is always an increase in the entropy of the universe (ΔSuniverse=ΔSsystem+ΔSsurroundingsΔSuniverse​=ΔSsystem​+ΔSsurroundings​).

3.

Second Law of Thermodynamics

d)

the different ways particles can order themselves

4.

Spatial Arrangement

e)

Entropy values for common substances at 25°C; perfect crystals at absolute zero have the lowest possible entropy values.

5.

S°

2.

Entropy can be described as

a)

the dispersal of matter

b)

the dispersal of energy

c)

the degree of disorder

d)

all of these

3.

Systems in nature tend to undergo changes toward

a)

lower energy and less disorder

b)

lower energy and more disorder

c)

higher energy and less disorder

d)

higher energy and more disorder

4.
Spontaneity is determined by....
a)
enthalpy only
b)
entropy only
c)
enthalpy and entropy
d)
enthalpy, entropy, and temperature
5.
Spontaneous reactions are driven by
a)
increasing enthalpy and increasing entropy.
b)
decreasing enthalpy and decreasing entropy.
c)
increasing enthalpy and decreasing entropy.
d)
decreasing enthalpy and increasing entropy.
6.

What are the best conditions to lead towards a spontaneous reaction?

a)

high negative enthalpy, high negative entropy

b)

high positive enthalpy, high positive entropy

c)

high negative enthalpy, high positive entropy

d)

high positive enthalpy, high negative entropy.

7.

Entropy increases from solid, liquid to gas. Why? (Multiple Answers)

a)

Molecular disorder increases

b)

Number of molecules increase in number from solid to gas

c)

As the phases change, they occupy more volume

d)

Molecules are heavier in solid; the density is greater

e)

The number of possible spatial arrangements increases as you go through phases

8.

How does increasing temperature affect entropy and why?

a)

Entropy decreases as kinetic energy decreases, so particles are more stable

b)

Entropy increases as kinetic energy increases, so the energy is more dispersed

c)

Entropy decreases as kinetic energy increases, so level of disorder decreases

d)

Entropy increases as kinetic energy decreases, so rate of reaction decreases

9.

The entropy will usually increase when

a)

a molecule is broken into two or more smaller molecules

b)

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

c)

a solid changes to a liquid

d)

all of these

10.
For the following reaction indicate if entropy is increased or decreased.     KClO3(s)  ------>  KCl(s)  +  O2(g) 
a)
increase
b)
decrease
11.
For the following reaction indicate if entropy is increased or decreased.    2N2(g)  +  3 H2(g)  ------->  2 NH3(g)
a)
increase
b)
decrease
12.

Predict the entropy change

a)

ΔS = +ve

b)

ΔS = -ve

c)

ΔS = 0

d)

no change in ΔS

13.

Predict the entropy change

a)

ΔS = +ve

b)

ΔS = -ve

c)

ΔS = 0

d)

no change in ΔS

14.
Predict the entropy change, ΔS.
a)
ΔS = +ve
b)
ΔS = -ve
c)
ΔS = 0
d)
ΔS = no change
15.
Predict the entropy change, ΔS.
a)
ΔS = +ve
b)
ΔS = -ve
c)
ΔS = 0
d)
ΔS = no change
16.

Which combination of ΔH and ΔS NEVER has a spontaneous reaction?

a)

+ΔH and +ΔS

b)

+ΔH and -ΔS

c)

-ΔH and -ΔS

d)

-ΔH and +ΔS

17.

Which factor affects the entropy of a system?

a)

Volume

b)

Temperature

c)

Pressure

d)

All of the above

18.

What does the entropy of a perfect crystal at absolute zero temperature tend to be according to the Third Law of Thermodynamics?

a)

Zero

b)

Infinite

c)

Constant

d)

Unpredictable

19.

Which reaction has a +ΔS ?

a)

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

b)

H2O(g) + CO2(g) → H2CO3(aq)

c)

H2(g) + I2(g) → 2Hl(g)

d)

C2H2O2(g) → 2CO(g) + H2(g)

20.

For a reaction to be spontaneous under standard conditions at all temperatures, the signs of ΔHΟ and ΔSΟ must be __________ and __________, respectively.

a)

+, +

b)

+, -

c)

-, +

d)

-, -

e)

+, 0

21.

The reaction 2 H2(g)+O2(g)→ 2 H2O(g)2\ H_{2\left(g\right)}+O_{2\left(g\right)}\rightarrow\ 2\ H_2O_{\left(g\right)}  shows 

a)

no change in entropy

b)

an increase in entropy

c)

a decrease in entropy

22.

Which of these would ALWAYS indicate a process is spontaneous?

a)

ΔH is + ΔS is +\Delta H\ is\ +\ \Delta S\ is\ +  

b)

ΔH is − ΔS is −\Delta H\ is\ -\ \Delta S\ is\ -  

c)

ΔH is + ΔS is −\Delta H\ is\ +\ \Delta S\ is\ -  

d)

ΔH is − ΔS is +\Delta H\ is\ -\ \Delta S\ is\ +  

23.

What is the correct expression for calculating ΔSsystem\Delta S_{system} ?

a)

ΣΔSreactnts −ΣΔSproductsΣ\Delta S_{reactnts\ }-Σ\Delta S_{products}

b)

ΣΔSproducts −ΣΔSreactntsΣ\Delta S_{products\ }-Σ\Delta S_{reactnts}

c)

ΣΔSproducts +ΣΔSreactntsΣ\Delta S_{products\ }+Σ\Delta S_{reactnts}

d)

ΣΔSreactnts +ΣΔSproductsΣ\Delta S_{reactnts\ }+Σ\Delta S_{products}

24.

Calculate the standard entropy change (ΔS sys) for the industrial synthesis of urea (a common fertilizer):
CO2(g) + 2 NH3(g) → CO(NH2)2(s) + H2O(ℓ)
at 25◦C the thermodynamic data shown is applicable. 
YES Calculator!

a)

−424.13 J/mol⋅K

b)

−212.07 J/mol⋅K

c)

−848.26 J/mol⋅K

d)

−1696.52 J/mol⋅K

25.

What is the correct expression for calculating ΔSsurroundings\Delta S_{surroundings} ?

a)

ΔSsurr = ΔH +T\Delta S_{surr\ }=\ \Delta H\ +T

b)

ΔSsurr = ΔHT\Delta S_{surr\ }=\ \frac{\Delta H}{T}

c)

ΔSsurr =− qsurrT\Delta S_{surr\ }=-\ \frac{q_{surr}}{T}

d)

ΔSsurr =− qsysT\Delta S_{surr\ }=-\ \frac{q_{sys}}{T}

26.

Calculate the entropy change of the surroundings (ΔSsurr) when one mole of water freezes at 0°C and a pressure of one atmosphere.

H2O(l) → H2O(s)
ΔH° = -6.02 kJ/mol

a)

  • 22.1 J/K

b)

  • 397 J/K

c)

  • -22.1 J/K

d)

  • -397 J/K

27.

What does Gibbs Free Energy tell us?

a)

How much energy is given off by a reaction.

b)

The tendency of a reaction to become "random"

c)

How spontaneous a reaction is.

28.

What information do we need to perform a calculation of Gibbs Free Energy?

a)

Enthalpy of the reaction.

b)

Entropy of the reaction.

c)

The temperature of the reaction in Kelvin.

d)

All of the options are needed.

29.

Enthalpy is a driving force when:

a)

exothermic

b)

endothermic

c)

there is an increase in entropy

d)

there is a decrease in entropy

30.

What is the equation for Gibbs Free Energy Change?

a)

ΔG = ΔH + TΔS

b)

ΔH = ΔG - TΔS

c)

ΔS = ΔH - TΔG

d)

ΔG = ΔH - TΔS

31.

How does Gibbs free energy change tell us a reaction is feasible?

a)

When G < 0

b)

When G > 0

32.

A reaction that has a –ΔH and a –ΔS is

a)

favorable at all temperatures

b)

favorable at high temperatures
i.e. when ΔH < TΔS

c)

unfavorable at all temperatures

d)

favorable at low temperatures
i.e. when ΔH > TΔS

33.

A reaction that has a –ΔH and a +ΔS is

a)

favorable at all temperatures

b)

favorable at high temperatures
i.e. when ΔH < TΔS

c)

unfavorable at all temperatures

d)

favorable at low temperatures
i.e. when ΔH >TΔS

34.

A reaction that has a +ΔH and a +ΔS is

a)

favorable at all temperatures

b)

favorable at high temperatures
i.e. when ΔH < TΔS

c)

unfavorable at all temperatures

d)

favorable at low temperatures
i.e. when ΔH >TΔS

35.

A reaction that has a +ΔH and a –ΔS is

a)

favorable at all temperatures

b)

favorable at high temperatures

i.e. when ΔH < TΔS

c)

unfavorable at all temperatures

d)

favorable at low temperatures
i.e. when ΔH > TΔS

36.

1. Given the following information, calculate ΔG0 for the reaction below at 250C.

SnCl4(l) + 2 H20(l) → SnO2(s) + 4HCl(g)

ΔH = 133.0 kJ and
ΔS = 401.5 J/K

a)

-252.6 kJ

b)

-13.4 kJ

c)

13.4 kJ

d)

252.6 kJ

37.

Your enthalpy is +20 and your entropy is +5.
At what temperature will the reaction be spontaneous?

a)

0 K

b)

4 K

c)

8 K

38.

Your enthalpy is -20 and your entropy is -8.

Which temperature will give you a spontaneous reaction?

a)

1 K

b)

3 K

c)

5 K

39.
A reaction has a positive ΔH and a positive ΔS. Which of the following is true?
a)
It will be spontaneous at all temperatures.
b)
It will be nonspontaneous at all temperatures.
c)

It will be spontaneous when ΔH > TΔS

d)

It will be spontaneous when ΔH < TΔS

40.

For the process at 250C : I2(g) →I2(s).

What are the signs for ΔG, ΔH and ΔS?

ΔH = ​ (a)  

ΔS = ​ (b)  

ΔG = ​ (c)  

Choose from the below words
-
- (at low T)
+
+
- (at high T)
41.

Predict the signs of ΔH , ΔS and ΔG for the reaction:

Mg(s) --> Mg(g)

ΔH =​ (a)  

ΔS = ​ (b)  

ΔG =​ ​ (c)  

Choose from the below words
+
- (if T is high)
-
- (if T is low)
-
42.

True/False: The speed of a chemical reaction is unrelated to the spontaneity of the reaction.

a)
True
b)
False
43.
Spontaneous reactions may be extremely slow.
a)
True
b)
False
44.

The formation
½ A2 + 2 B2 + C --> CAB4

has an enthalpy of formation of -104 kJ and a change in entropy of -60.8 J/K at 30 °C.
What is the free energy and spontaneity of the reaction?

a)

-85.6 kJ, spontaneous

b)

-18.3 kJ, not spontaneous

c)

+18.3 kJ, spontaneous

d)

+85.6 kJ, not spontaneous

45.

What is the standard unit for Gibbs free energy change?

a)

kJ/mol

b)

N/m

c)

J/mol

d)

mol/dm³

46.

At what temperature would a given reaction become spontaneous if ΔH =+119kJ and ΔS=+263J/K.mol

a)

455 K

b)

-2210 K

c)

382 K

d)

-363 K

47.

Free energy change at equilibrium is...

a)

Zero

b)

Positive

c)

Negative

48.
A 100mL sample of water is placed in a coffee cup calorimeter.  Solid NaCl is then dissolved in the water.  The temperature of the water decreases from 20.5°C to 19.7°C and is then allowed to return to room temperature (20.5°C).  Determine the signs for ΔH and ΔS for the process of dissolving NaCl and 
ΔG for the entire process at constant temperature.
NO Calculator!
a)

+ΔH, −ΔS, ΔG=0

b)

−H, +S, ΔG=0

c)

+ΔH, +ΔS, −ΔG

d)

−ΔH, −ΔS, −ΔG

49.

2C(s) + 2H2(g) → C2H4(g) ΔH0rxn = +52.4 kJ/molrxn

It is observed that the reaction producing C2H4(g) from its elements goes essentially to completion. Which of the following is a true statement about the thermodynamic favorability of the reaction?

a)

The reaction is unfavorable due to both enthalpy and entropy changes.

b)

The reaction is favorable and driven by an entropy change only.

c)

The reaction is favorable and driven by an enthalpy change only.

d)

The reaction is favorable and driven by both enthalpy and entropy changes.

50.

Which of the following sets of conditions must be true for a reaction that is thermodynamically favorable only at relatively high temperatures?

a)

-ΔG

-ΔH

-ΔS

b)

-ΔG

+ΔH

+ΔS

c)

+ΔG

-ΔH

-ΔS

d)

+ΔG

+ΔH

+ΔS

51.

If a reaction increases the # of moles of gas, then the sign for ΔS is what?

a)

Positive

b)

Negative

c)

Neutral

d)

Equilibrium

52.

Calculating ΔG for melting ice:


ΔHFus= 6.01 kJ/mol-rxn

ΔSFus= 22.0 kJ/mol-rxn*K

20oC (K=273+co)


ΔG=ΔH-TΔS

a)

0.44 kJ/mol-rxn, spontaneous

b)

-0.44 kJ/mol-rxn, nonspontaneous

c)

0.44 kJ/mol-rxn, nonspontaneous

d)

-0.44 kJ/mol-rxn, spontaneous

53.

The First Law of Thermodynamics indicates that

a)

Work is force exerted over a distance.

b)

Energy in a universe is finite.

c)

An increase in disorder is part of any spontaneous process.

d)

Work is a state function.

54.

The Third Law of Thermodynamics indicates that the

a)

enthalpy of the universe is constant.

b)

entropy of the universe increases.

c)

sum of the entropies of the system plus surroundings is zero.

d)

entropy of a perfectly ordered pure crystal is zero at absolute zero.

55.

The substance with the smallest value of So is

a)

AlCl3

b)

MgCl2

c)

NaCl

d)

SnCl4

56.

Entropy is always ________ in the Universe

a)

increasing

b)

decreasing

c)

constant

57.

What is a thermodynamically unfavorable reaction?

a)

A process where the entropy of the system decreases

b)

A reaction that occurs spontaneously without an external energy source

c)

A reaction in which ΔG°< 0

d)

A reaction that does not occur spontaneously without an external energy source

58.

a reaction that is exothermic and has a decrease in entropy is

a)

spontaneous at high temps

b)

spontaneous at low temps

c)

always spontaneous

d)

never spontaneous

59.

Which of the following would have a nonzero enthalpy of formation?

a)

Fe(s)

b)

Br2(s)

c)

O2(g)

d)

Hg(l)

60.

Which of the following equations represents a reaction with a negative ΔS°?

a)

2NO(g) + O2(g) → 2NO2(g)

b)

2KClO3(s) → 2KCl(s) + 3O2(g)

c)

PCl5(g) → PCl3(g) + Cl2(g)

d)

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

61.

The oxidation of PCl3(g) is represented by the equation above, and the table provides the approximate values of the absolute molar entropies, S°, for these substances. Based on the information given, what is the approximate ΔS° for the reaction?

a)

+170J/(molrxn⋅K)

b)

−170J/(molrxn⋅K)

c)

+190J/(molrxn⋅K)

d)

−190J/(molrxn⋅K)

62.

CH3OH(g) --> CO(g) + 2H2(g)

The reaction represented above goes essentially to completion. The reaction takes place in a rigid, insulated vessel that is initially at 600 K

What can be inferred about ∆S° for the reaction at 600 K?

a)

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

b)

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

c)

It must be positive, since ∆G° is negative and ∆H° is positive.

d)

It must be negative, since ∆G° is positive and ∆H° is positive.

63.

Reaction 1: 4 NH3(g) + 8O2(g) → 4 HNO3(aq) + 4 H2O(I)
Based on the values of ΔG° for the three reactions represented below, what is the value of ΔG° for Reaction 1 represented above?


Reaction 2: 4NH3(g) + 5O2(g) → 4 NO(g) + 6 H2O(l) ΔG° = -1010 kJ/mol

Reaction 3: 2NO2(g) → 2 NO(g) + O2(g) ΔG° = 70 kJ/mol

Reaction 4: 4NO2(g) + O2(g) + 2H2O(l) → 4 HNO3(aq) ΔG° = -170 kJ/mol

a)

-1040 kJ/mol

b)

-1110 kJ/mol

c)

-1250 kJ/mol

d)

-1320 kJ/mol

64.

Although the dissolution of ammonium chloride (NH4Cl) in water is an endothermic reaction, even then it is spontaneous because:

a)

  1. TΔS<ΔH

b)

  1. ΔS=−ve and TΔS<ΔH

c)

  1. ΔS=0

d)

  1. ΔS=−ve

65.

Which combination of acid (HCl) and base (KOH) will produce the greatest amount of heat?

a)

1M of 10mL HCl with

1M of 10mL KOH

b)

2M of 10mL HCl with

0.5M of 10mL KOH

c)

0.5M of 10mL HCl with

0.5M of 10mL KOH

d)

0.5M of 10mL HCl with

1M of 10mL KOH

66.

Which of the following methods can be used to make a nonspontaneous reaction (ΔG > 0) proceed? (TWO correct answers)

a)

Lowering the temperature

b)

Increasing the pressure

c)

Applying external energy

d)

Decreasing the concentration of reactants

e)

Pairing up the reaction with another (coupling)

67.

Consider the following reaction:

2 NO (g) + O₂ (g) → 2 NO₂ (g) ΔG° = +70 kJ/mol

Which of the following reactions, when coupled with the main reaction, can make the overall process of converting NO to NO₂ more thermodynamically favorable?

a)

2H₂O(g) → 2H₂ (g) + O₂ (g)

ΔG° ≈ -237 kJ/mol

b)

Cl₂ (g) + H₂O (l) → HCl (g) + HClO (aq)

ΔG° ≈ +254 kJ/mol

c)

2 NO₂ (g) + H₂O (l) → HNO₃ (aq) + HNO₂ (aq) ΔG° ≈ -140 kJ/mol

d)

CO (g) + 2 NO₂ (g) → CO₂ (g) + N₂O₅ (g)

ΔG° ≈ -137 kJ/mol


68.

What is the main purpose of Coupling Reactions?

a)

To create increase the free energy of the overall process

b)

To increase the activation energy of a reaction.

c)

To decrease the rate of a reaction.

d)

To make a thermodynamically unfavorable reaction proceed by utilizing a favorable one.

69.

6. The formation ½ A2 + 2 B2 + C --> CAB4 has an enthalpy of formation of -104 kJ and a change in entropy of -60.8 J/K at 30 °C. What is the free energy and spontaneity of the reaction?

a)

-85.6 kJ, spontaneous

b)

-18.3 kJ, not spontaneous

c)

+18.3 kJ, spontaneous

d)

+85.6 kJ, not spontaneous

70.

Which reaction has the greatest increase in entropy of the system?

A. HCl (g) + NH3 (g) → NH4Cl (s)

B. (NH4)2Cr2O7 (s) → Cr2O3 (s) + N2 (g) + 4H2O (g)

C. CaCO3 (s) → CaO (s) + CO2 (g)

D. I2 (g) → I2 (s)

a)

A

b)

B

c)

C

d)

D

71.
A reaction has a positive ΔH and a positive ΔS. Which of the following is true?
a)
It will be spontaneous at all temperatures.
b)
It will be nonspontaneous at all temperatures.
c)
It will be spontaneous at low temperatures.
d)
It will be spontaneous at high temperatures.
72.

What can be said about a chemical system that has reached a minimum in free energy?

a)

The system has achieved equilibrium.

b)

The reaction is complete.

c)

The system entropy is zero.

d)

The reaction is very fast.

73.

At what temperature would a given reaction become spontaneous if ΔH =+119kJ and ΔS=+263J/K.mol

a)

455 K

b)

-2210 K

c)

382 K

d)

-363 K

74.

For any reaction at equilibrium, which of the following is true?

a)

ΔH<0

b)

ΔS< 0

c)

ΔG=0

d)

ΔS=0

75.

Given the change of phase:

CO2(g) —> CO2(s)

Predict the change in entropy of the process above. Justify your answer.

Categorize the following

ΔS is negative

solid particles occupy less volume than gas

solid particles are less dispersed than gas

a gas is turning into a solid

solids have less possible spatial configurations than gases

the process is spontaneous

ΔS is positive

ΔS is zero

the kinetic energy of the solid particles is more dispersed

Claim (1 point)
Evidence/Reasoning (4 points)
Irrelevant/Incorrect
76.

The correct statements about entropy and gibbs free energy (choose 3 instead of 4)

a)

The entropy change is calculated in energy units of joules, but ΔG° and ΔH° are both measured in kJ.

b)

ΔG = ΔH – TΔS

c)

ΔG = ΔH + TΔS

d)

For a reaction to be feasible, ΔG has to be negative.

77.
The equation for Gibbs at STANDARD conditions (ΔG°) is....
a)
ΔG° = - RTlnK
b)
You're not actually going to click this.
c)
Seriously, you'd be crazy to click this.
d)
Click this (just kidding - don't).
78.

For the equation,
ΔG° = -RTlnK
If K is less than 1, what is the value of ΔG°?

a)

ΔG = 0

b)

ΔG is negative since the reaction is reactant favoured

c)

ΔG is positive since the reaction is product favoured

d)

ΔG is positive since the reaction is reactant favoured

79.

For the equation,
ΔG° = -RTlnK
If K is more than 1, (K>1) what is the value of ΔG°?

a)

ΔG = 0

b)

ΔG is positive since the reaction is reactant favoured

c)

ΔG is negative since the reaction is product favoured

d)

ΔG is negative since the reaction is reactant favoured

80.
The equation for Gibbs at NONSTANDARD conditions (ΔG) is....
a)
ΔG = ΔG° + RTlnQ
b)
Don't click this.
c)
Seriously, don't click this.
d)
I can't believe you clicked this.
81.
If the equilibrium constant for a reaction is LARGE, this means......
a)
ΔG° is very negative and spontaneous.
b)
ΔG° is very negative and nonspontaneous.
c)
ΔG° is very positive and spontaneous.
d)
ΔG° is very positive and nonspontaneous.
82.
If ΔG becomes more negative compared to ΔG°.....
a)
that means the reaction is less spontaneous
b)
that means the reaction is more spontaneous
83.
For the equation,
ΔG = ΔG° + RTlnQ
if Q > K
a)
ΔG increases (less spontaneous)
b)
ΔG decreases (more spontaneous)
c)
ΔG increases (more spontaneous)
d)
ΔG decreases (less spontaneous)
84.

The graph shows Gibbs free energy of a mixture of N2O4 (g) and NO2 (g) in different proportions. Which point shows the system at equilibrium?

N2O4 (g) ⇌ 2NO2 (g)

a)

A

b)

B

c)

C

d)

D

85.

Which statement is correct for a spontaneous reaction?

a)

A

b)

B

c)

C

d)

D

86.

Which corresponds to a system at equilibrium?

a)

A

b)

B

c)

C

d)

D

87.

What is the limitation of using ΔG to predict the feasibility of a reaction?

a)

ΔG can only be used to predict the feasibility of a reaction under standard conditions.

b)

ΔG indicates that the reaction will occur at an observable rate.

c)

ΔG takes into account the kinetics of the reaction.

d)

ΔG can predict the exact time when the reaction will occur.

88.

If the equilibrium constant (K) for a reaction is greater than 1, what is true about the reaction?

a)

Reactants are favored.

b)

Neither reactants nor products are favored.

c)

Products are favored.

d)

The reaction is at equilibrium.

89.

Rearrange the Gibbs free energy equation ( ΔG = ΔH - TΔS) to solve for the temperature at a phase transition e.g. H2O (l) --> H2O (g)

a)

T=ΔH−ΔGΔST=\frac{ΔH-ΔG}{ΔS}

b)

T=ΔHΔST=\frac{ΔH}{ΔS}

c)

T = - ΔG

d)

T=−ΔGΔST=\frac{-ΔG}{ΔS}

90.

The equation for Gibbs free energy in STANDARD conditions (ΔG°) is....

a)
ΔG° = - RTlnK
b)
You're not actually going to click this.
c)
Seriously, you'd be crazy to click this.
d)
Click this (just kidding - don't).
91.

What is the correct rearrangement for calculating K from ΔG?

ΔG = -RTln(K)

a)

K=eΔGRTK=e^{\frac{ΔG}{RT}}

b)

K=eΔG⋅RTK=e^{ΔG\cdot RT}

c)

K=e−ΔGRTK=e^{\frac{-ΔG}{RT}}

d)

K=eΔG+RTK=e^{ΔG+RT}

92.

The graph shows Gibbs free energy of a mixture of N2O4 (g) and NO2 (g) in different proportions. Which point shows the system at equilibrium?

N2O4 (g) ⇌ 2NO2 (g)

a)

A

b)

B

c)

C

d)

D

93.

Which statement is correct for a spontaneous reaction?

a)

A

b)

B

c)

C

d)

D

94.

Which corresponds to a system at equilibrium?

a)

A

b)

B

c)

C

d)

D

95.

If the equilibrium constant (K) for a reaction is greater than 1, what is true about the reaction?

a)

Reactants are favored.

b)

Neither reactants nor products are favored.

c)

Products are favored.

d)

The reaction is at equilibrium.

96.
If the equilibrium constant for a reaction is LARGE, this means......
a)
ΔG° is very negative and spontaneous.
b)
ΔG° is very negative and nonspontaneous.
c)
ΔG° is very positive and spontaneous.
d)
ΔG° is very positive and nonspontaneous.
97.

H2(g) + I2(g) ⇋ 2HI(g)
For the reaction above the standard free energy is +1.7 kJ/mol

Therefore, what would the value of the equilibrium constant be?

a)
K = 0
b)
K = 1
c)
K > 1
d)
K < 1
98.

For the equation,
ΔG° = -RTlnK
If K is less than 1, what is the value of ΔG°?

a)

ΔG = 0

b)

ΔG is negative since the reaction is reactant favoured

c)

ΔG is positive since the reaction is product favoured

d)

ΔG is positive since the reaction is reactant favoured

99.

For the equation,
ΔG° = -RTlnK
If K is more than 1, (K>1) what is the value of ΔG°?

a)

ΔG = 0

b)

ΔG is positive since the reaction is reactant favoured

c)

ΔG is negative since the reaction is product favoured

d)

ΔG is negative since the reaction is reactant favoured

100.

X(g) + Y(g) ⇌ XY(g)

In an experiment, the reaction above reaches equilibrium at 300s.
What can be concluded about the sign of ΔG for the reaction from 0s until 300s?.

a)

We cannot determine the sign of ΔG

b)

ΔG is negative as written, since the system shifts to the products

c)

ΔG is positive as written, since the system shifts to the products

d)

ΔG is negative as written, since the system shifts to the reactants

e)

ΔG is positive as written, since the system shifts to the reactants

101.

Calculate the equilibrium constant for the reaction below at 25oC:

N2O4(g) → 2NO2(g) ΔG°rxn = +4.8 kJ/mol

ΔG°rxn = - RT InK

a)

0.14

b)

1.14

c)

1.40

d)

0.40

102.

What is the limitation of using ΔG to predict the feasibility of a reaction?

a)

ΔG can only be used to predict the feasibility of a reaction under standard conditions.

b)

ΔG indicates that the reaction will occur at an observable rate.

c)

ΔG takes into account the kinetics of the reaction.

d)

ΔG can predict the exact time when the reaction will occur.

103.
The equation for Gibbs at NONSTANDARD conditions (ΔG) is....
a)
ΔG = ΔG° + RTlnQ
b)
Don't click this.
c)
Seriously, don't click this.
d)
I can't believe you clicked this.
104.
For the equation,
ΔG = ΔG° + RTlnQ
if Q > K
a)
ΔG increases (less spontaneous)
b)
ΔG decreases (more spontaneous)
c)
ΔG increases (more spontaneous)
d)
ΔG decreases (less spontaneous)
105.
If ΔG becomes more negative compared to ΔG°.....
a)
that means the reaction is less spontaneous
b)
that means the reaction is more spontaneous
106.

Which expression can be used to plot a useful graph for finding ΔH° from the following equation?
ΔG°= –RT ln(K) = ΔH°–TΔS°

a)

ln⁡(K)=−ΔHoR(1T)+ΔSoR\ln\left(K\right)=-\frac{\Delta H^o}{R}\left(\frac{1}{T}\right)+\frac{\Delta S^o}{R}

b)

ln⁡(K)=ΔHoRT−ΔSoR\ln\left(K\right)=\frac{\Delta H^o}{RT}-\frac{\Delta S^o}{R}

c)

−ΔHoR(1T+ΔSoR)ln⁡K-\frac{\Delta H^o}{R}\left(\frac{1}{T}+\frac{\Delta S^o}{R}\right)\ln K

107.

What is the value of the slope?
ΔG°= –RT ln(K) = ΔH°–TΔS°

ln⁡(K)=−ΔHoR(1T)+ΔSoR\ln\left(K\right)=-\frac{\Delta H^o}{R}\left(\frac{1}{T}\right)+\frac{\Delta S^o}{R}

a)


slope = −ΔSRslope\ =\ \frac{-ΔS}{R}

b)


slope = −ΔHRslope\ =\ \frac{-ΔH}{R}

c)


slope = ΔHRslope\ =\ \frac{ΔH}{R}

d)


slope = ΔSRslope\ =\ \frac{ΔS}{R}

108.

What is the value of the y intercept?

ΔG°= –RT ln(K) = ΔH°–TΔS°

ln⁡(K)=−ΔHoR(1T)+ΔSoR\ln\left(K\right)=-\frac{\Delta H^o}{R}\left(\frac{1}{T}\right)+\frac{\Delta S^o}{R}

a)


y intercept = −ΔSRy\ intercept\ =\ \frac{-ΔS}{R}

b)


y intercept = −ΔHRy\ intercept\ =\ \frac{-ΔH}{R}

c)


y intercept = ΔHRy\ intercept\ =\ \frac{ΔH}{R}

d)


y intercept = ΔSRy\ intercept\ =\ \frac{ΔS}{R}

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