Thermodynamics Test

Result

Thermodynamics Test - 25

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following conditions guarantee a spontaneous reaction?

A
Positive $$\triangle H$$, positive $$\triangle S$$

B
Positive $$\triangle H$$, negative $$\triangle S$$

C
Negative $$\triangle H$$, negative $$\triangle S$$

D
Negative $$\triangle H$$, positive $$\triangle S$$

E
None of the above

Solution

Spontaneity of a reaction can be given by the equation $$\Delta G$$$$=$$$$\Delta H$$$$-$$$$T$$$$\Delta S$$. If $$\Delta G$$ is negative, the reaction can be said to be spontaneous.
If $$\Delta S$$ has a positive value, $$T$$$$\Delta S$$ will be negative in nature because of a negative sign in front. Again if $$\Delta H$$ is negative in nature, the entire value of the eqiuation will be negative in nature. Thus $$\Delta G$$ will also be negative. Hence reaction is spontaneous.
Option D
Question 2
1 / -0

The heat capacity for aluminum is $$\displaystyle 0.89{ Jg }^{ -10 }{ C }^{ -1 }$$, for iron is $$\displaystyle 0.45{ Jg }^{ -10 }{ C }^{ -1 }$$, and for zinc is $$\displaystyle 0.39{ Jg }^{ -10 }{ C }^{ -1 }$$. If 100 J of heat energy was added to a10 g sample of each of the metals, which of the following would be true?

A
$$\displaystyle Al$$ would have the largest temperature increase.

B
$$\displaystyle Fe$$ would have the largest temperature increase.

C
$$\displaystyle Zn $$ would have the largest temperature increase.

D
$$\displaystyle Fe$$ and $$\displaystyle Zn$$ would have the same increase.

E
All three metals would have the same temperature increase.

Solution

The heat capacity is expressed by the formula $$s$$$$=$$$$Q$$$$/$$$$m$$$$\Delta T$$, where $$s$$ is the specific heat capacity, $$m$$ is the mass of sample, $$Q$$ is the amount of heat energy added and $$\Delta T$$ is the change in temperature. Putting all the values in the given equation, we will get,
For Zinc, $$s$$ is $$0.39$$, $$Q$$ is $$100$$, $$m$$ is $$10$$ and $$\Delta T$$ is unknown. Hence-
$$0.39$$$$=$$$$100$$$$/$$$$10$$$$\Delta T$$. From this, the value of $$\Delta T$$ has been to be found as $$25.64$$.
For Aluminium and Iron, the change in temperature if calculated will be found to be lesser than Zinc.
Hence, Zinc would have the largest temperature increase.
Note- Here the option chosen as answer has some printing mistake. It should be $$Zn$$.
Question 3
1 / -0

To which portion of the heating curve for water does $$\Delta {H}_{vap}$$ apply?

A
$$A-B$$

B
$$B-C$$

C
$$C-D$$

D
$$D-E$$
Solution
- From the heating curve of water ,DE is the region where both water and vapor simultaneously occur.
- So DE is the region where $$\Delta H_{vapor}$$ can be applied.
Question 4
1 / -0

$$C_v$$ for helium gas $$(He)$$ is $$(in J \ mol^{-1} \ K^{-1})$$

A
2.5

B
10.5

C
1.5

D
12.5

Solution

The internal energy of 1 mole of a gas at temperature $$T$$ , having $$f$$ degrees of freedom is given by ,
$$U=N_{A}\times(1/2)fk_{B}T=(1/2)fRT$$ ......................eq1
where $$N_{A}=$$Avogadro's number
$$k_{B}=$$ Boltzmann's constant
from first law of thermodynamics ,
$$dU=dQ-dW$$ ............eq2
At constant volume , $$dV=0$$
hence $$dW=PdV=0$$
eq2 becomes , $$dU=dQ$$
but $$dQ=C_{V}dT$$ (for 1 mole of gas)
therefore $$dU=C_{V}dT$$
or $$C_{V}=dU/dT$$
putting the value of $$U$$ in this equation, we get
$$C_{V}=\frac{d(1/2)fRT}{dT}=\frac{f}{2}R$$
for Helium , $$f=3$$ as it is a monatomic gas ,
hence $$C_{V}=3R/2$$
we have $$R=8.31Jmol^{-1}K^{-1}$$
Therefore $$C_{V}=\frac{3\times8.31}{2}=12.5Jmol^{-1}K^{-1}$$

Question 5

1 / -0

Consider a spontaneous electrochemical cell between $$Cu$$ and $$Al$$. Predict what would happen if excess concentrated $$NaOH$$ were added to the cell with copper ions and a precipitate forms.

Standard Potential (V)	Reduction Half-Reaction
$$2.87$$	$$F_{2}(g) + 2e^{-} \rightarrow 2F^{-}(aq)$$
$$1.51$$	$$MnO_{4}^{-}(aq) + 8H^{+}(aq) + 5e^{-}\rightarrow Mn^{2+}(aq) + 4H_{2}O(l)$$
$$1.36$$	$$Cl_{2}(aq) + 3e^{-} \rightarrow 2Cl^{-}(aq)$$
$$1.33$$	$$Cr_{2}O_{7}^{2-} (aq) + 14H^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3+}(aq) + 7H_{2}O(l)$$
$$1.23$$	$$O_{2}(g) + 4H^{+}(aq) + 4e^{-}\rightarrow 2H_{2}O(l)$$
$$1.06$$	$$Br_{2}(l) + 2e^{-} \rightarrow 2Br^{-}(aq)$$
$$0.96$$	$$NO_{3}^{-}(aq) + 4H^{+}(aq) + 3e^{-}\rightarrow NO(g) + H_{2}O(l)$$
$$0.80$$	$$Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$$$
$$0.77$$	$$Fe^{3+} (aq) + e^{-} \rightarrow Fe^{2+}(aq)$$
$$0.68$$	$$O_{2}(g) + 2H^{+}(aq) + 2e^{-}\rightarrow H_{2}O_{2}(aq)$$
$$0.59$$	$$MnO_{4}^{-}(aq) + 2H_{2}O(l) + 3e^{-}\rightarrow MnO_{2}(s) + 4OH^{-}(aq)$$
$$0.54$$	$$I_{2}(s) + 2e^{-}\rightarrow 2I^{-}(aq)$$
$$0.40$$	$$O_{2}(g) + 2H_{2}O(l) + 4e^{-} \rightarrow 4OH^{-}(aq)$$
$$0.34$$	$$Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$$
$$0$$	$$2H^{+}(aq) + 2e^{-}\rightarrow H_{2}(g)$$
$$-0.28$$	$$Ni^{2+}(aq) + 2e^{-}\rightarrow Ni(s)$$
$$-0.44$$	$$Fe^{2+}(aq) + 2e^{-}\rightarrow Fe(s)$$
$$-0.76$$	$$Zn^{2+}(aq) + 2e^{-}\rightarrow Zn(s)$$
$$-0.83$$	$$2H_{2}O(l) + 2e^{-}\rightarrow H_{2}(g) + 2OH^{-}(aq)$$
$$1.66$$	$$Al^{3+}(aq) + 3e^{-}\rightarrow Al(s)$$
$$-2.71$$	$$Na^{+}(aq) + e^{-} \rightarrow Na(s)$$
$$-3.05$$	$$Li^{+}(aq) + e^{-}\rightarrow Li(s)$$

Voltage would increase since more ions can give up electrons.

Voltage will decrease since there will be less $$Cu^{2+}$$ ions in solution.

Voltage would stay the same since $$NaOH$$ is a strong electrolyte and will exist as a spectator ion.

Voltage would increase and then decrease as the addition of new ions reach equilibrium.

Solution

$$\text{More number of ions tends to flow more current so Voltage will be increases.}$$

Question 6
1 / -0

ln which of the processes, does the internal energy of the system remain constant?

A
Adiabatic

B
Isochoric

C
Isobaric

D
Isothermal

Solution

The change in internal energy of a system is given by $$\Delta U = C_V\Delta T$$
Thus, if the temperature of the system changes, then the internal energy changes.
Hence, internal energy remains constant for an "Isothermal" process as temperature remains constant in isothermal process.
Question 7
1 / -0

Given, sublimation and ionization energy of $$Na$$ are $$107\space kJ/mol$$ and $$502 \space kJ/mol$$ respectively and bond dissociation energy required for chlorine gas and its electron affinity energy are $$121\space kJ/mol$$ and $$-355\space kJ/mol$$. If $$\triangle H_{f}^{0}$$ is $$-411\space kJ/mol$$. What is its approximate lattice enthalpy?

A
$$-786\space kJ/mol$$

B
$$788\space kJ/mol$$

C
$$388\space kJ/mol$$

D
$$-508\space kJ/mol$$

Solution

$$\Delta H_{f^0} = \Delta H_{sub} + IE+\Delta H_{diss} + EA + U$$
$$\Delta H_{f^0} = 108+496+122-349-788 = 411\, kJ/mol$$
The enthalpy change in the formation of an ionic lattice from the gaseous isolated sodium and chloride ions is $$-788$$ $${ kJ }/{ mole }$$. That enthalpy change, which corresponds to the reaction $${ Na }_{ (g) }+{ Cl }_{ (g) }\rightarrow { NaCl }_{ (s) }$$, is called the lattice energy of the ionic crystal. Although the lattice energy is not directly measurable, there are various ways to estimate it from theoretical considerations and some experimental values. For all known ionic crystals, the lattice energy has a large negative value. It is ultimately the lattice energy of an ionic crystal which is responsible for the formation and stability of ionic crystal structures.
For sodium chloride, the Born - Haber cycle is as shown in the image.
Question 8
1 / -0

Correct mathematical formulation of first law of thermodynamics used in thermochemistry is :

A
$$\Delta H =\ \Delta U +\ \Delta n RT$$

B
$$\Delta U =\ \Delta H +\ \Delta n RT$$

C
$$\Delta H =\ \Delta U -\ \Delta n RT$$

D
$$\Delta H =\ -\Delta U +\ \Delta n RT$$

Solution

The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. Mathematically, this is represented as
$$\Delta H=\Delta U+\Delta nRT$$
with,
$$\Delta H$$ is the heat exchanged between a system and its surroundings,
$$\Delta U$$ is the total change in internal energy of a system,
$$\Delta nRT$$ is the work done by or on the system.
Question 9
1 / -0

To which portion of the heating curve for water does $$\Delta {H}_{fus}$$ apply?

A
$$A-B$$

B
$$B-C$$

C
$$C-D$$

D
$$D-E$$

Solution

From the heating curve of water in the region BC both ice and water exist simultaneously.So BC is the region where $$\Delta H_{fus}$$ can be applied.
Question 10
1 / -0

Based on knowledge of solid compounds and bond strengths, which of the following reactions is considered non-spontaneous due to the extremely high activation energy but once started becomes extremely spontaneous?

A
Dissolution of sodium hydroxide.

B
Creating a dilution of hydrochloric acid.

C
The burning of $$C_{20}H_{42}$$.

D
The synthesis of silver oxide.

Solution

$$\text{Dissolution of NaOH required high activation energy but once started becomes extremely spontaneous.}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 25

Result

Thermodynamics Test - 25

Score

-

Rank

-

SHARING IS CARING

Question 1

Positive $$\triangle H$$, positive $$\triangle S$$

Positive $$\triangle H$$, negative $$\triangle S$$

Negative $$\triangle H$$, negative $$\triangle S$$

Negative $$\triangle H$$, positive $$\triangle S$$

None of the above

Solution

Question 2

$$\displaystyle Al$$ would have the largest temperature increase.

$$\displaystyle Fe$$ would have the largest temperature increase.

$$\displaystyle Zn $$ would have the largest temperature increase.

$$\displaystyle Fe$$ and $$\displaystyle Zn$$ would have the same increase.

All three metals would have the same temperature increase.

Solution

Question 3

$$A-B$$

$$B-C$$

$$C-D$$

$$D-E$$

Solution

Question 4

2.5

10.5

1.5

12.5

Solution

Question 5

Voltage would increase since more ions can give up electrons.

Voltage will decrease since there will be less $$Cu^{2+}$$ ions in solution.

Voltage would stay the same since $$NaOH$$ is a strong electrolyte and will exist as a spectator ion.

Voltage would increase and then decrease as the addition of new ions reach equilibrium.

Solution

Question 6

Adiabatic

Isochoric

Isobaric

Isothermal

Solution

Question 7

$$-786\space kJ/mol$$

$$788\space kJ/mol$$

$$388\space kJ/mol$$

$$-508\space kJ/mol$$

Solution

Question 8

$$\Delta H =\ \Delta U +\ \Delta n RT$$

$$\Delta U =\ \Delta H +\ \Delta n RT$$

$$\Delta H =\ \Delta U -\ \Delta n RT$$

$$\Delta H =\ -\Delta U +\ \Delta n RT$$

Solution

Question 9

$$A-B$$

$$B-C$$

$$C-D$$

$$D-E$$

Solution

Question 10

Dissolution of sodium hydroxide.

Creating a dilution of hydrochloric acid.

The burning of $$C_{20}H_{42}$$.

The synthesis of silver oxide.

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.