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.

MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 4

Two substances P and Q when brought together, form substance R with the evolution of heat. The properties of R are different from both P and Q. What is substance R?

0%
A compound
0%
An element
0%
A metal
0%
A mixture

Explanation

Since the substance

$R$ is a mixture of two substances, it may either be a mixture or a compound

In a mixture the constituents retain their properties while in a compound, they lose their properties.

Since the question mentions that

$R$ has different property from both

$P$ and

$Q$ , hence it must be a compound

Which describes a nonspontaneous reaction?

0%
Positive $\displaystyle \Delta H$
0%
Negative $\displaystyle \Delta H$
0%
Positive $\displaystyle \Delta G$
0%
Positive $\displaystyle \Delta S$
0%
None of these

Which is the amount of energy that must be added to raise the temperature of 1 gram of a substance by

$\displaystyle { 1 }^{ \circ }C$ ?

0%
Enthalpy change
0%
Entropy change
0%
Gibbs free energy change
0%
Activation energy
0%
Specific heat capacity

Which portion of the heating curve for water shown would there be ONLY liquid water present?

0%
$A-B$
0%
$B-C$
0%
$C-D$
0%
$D-E$

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?

0%
$\displaystyle Al$ would have the largest temperature increase.
0%
$\displaystyle Fe$ would have the largest temperature increase.
0%
$\displaystyle Zn$ would have the largest temperature increase.
0%
$\displaystyle Fe$ and $\displaystyle Zn$ would have the same increase.
0%
All three metals would have the same temperature increase.

Explanation

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

$Zn$ .

Which of the following condition guarantee a spontaneous reaction?

0%
Positive $\displaystyle \Delta H$ , Positive $\displaystyle \Delta S$
0%
Positive $\displaystyle \Delta H$ , Negative $\displaystyle \Delta S$
0%
Negative $\displaystyle \Delta H$ , Negative $\displaystyle \Delta S$
0%
Negative $\displaystyle \Delta H$ , Positive $\displaystyle \Delta S$
0%
None of the above

Identify the equation used to determine the amount of heat required to melt 10 grams of ice.

0%
$Q = mC_{sp} \triangle T$
0%
$Q = n\triangle H$
0%
$KE = \dfrac{1}{2} mv^2$
0%
$PE = mgh$
0%
$PV = nRT$

Explanation

Option A is the correct answer.

To calculate the amount of heat required to melt a piece of ice, we need to know the mass of ice (

$m$ ), the change in temperature (

$\Delta T$ ), as we are raising the temperature. We also need to know the specific heat (

$C_{sp}$ ). Thus the equation can be written as:

$Q$

$=$

$m$

$C_{sp}$

$\Delta T$

Which of the following is technique used to measure the heat of a reaction?

0%
Gibbs Free Energy
0%
Entropy
0%
Enthalpy
0%
Calorimetry

$I$ . The energy content of reactants in a chemical reaction must be equal to the energy content of the products.

$II$ . Energy is conserved in chemical processes.

0%
Statement $I$ is true, Statement $II$ is true and is a correct explanation of the phenomena described in $I$
0%
Statement $I$ is true, Statement $II$ is false
0%
Statement $I$ is false, Statement $II$ is true
0%
Statement $I$ is false, Statement $II$ is false

State True or False:

I: Candles can be safely stored at room temperature/ even though their reaction with air is spontaneous at room temperature.
The reaction that takes place when a candle is burned involves a decrease in entropy.

0%
True
0%
False

How many calories are needed to heat

$100\ g$ of water from

$278\ K$ to

$288\ K$ ?

0%
$10$ calories
0%
$60$ calories
0%
$100$ calories
0%
$1000$ calories

Explanation

$Q=nC\triangle T$

$n=100g$

$C=$ specific heat capacity of water=

$1Jg^{-1}{^{o}C}^{-1}$

$\triangle T=(288K-278K)$

$\triangle T=10$

$Q=100\times 1\times10$

$Q=1000$ calories.

Statement I: A reaction with a positive enthalpy and a negative entropy will be spontaneous
BECAUSE
Statement II: The Gibbs free energy for a spontaneous reaction is negative

0%
Statement $1$ and Statement $2$ are correct and Statement 2 is the correct explanation of Statement $1$
0%
Both the Statement $1$ and Statement $2$ are correct and Statement $2$ is NOT the correct explanation of Statement $1$
0%
Statement $1$ is correct but Statement $2$ is not correct
0%
Statement $1$ is not correct but Statement $2$ is correct
0%
Both the Statement $1$ and Statement $2$ are not correct

Explanation

We know the formula

$\Delta G$

$=$

$\Delta H$

$-$

$T$

$\Delta S$

Positive enthalpy means

$\Delta H$ is

$+ve$ .

The second term on the right hand side of the equation is already

$-ve$ . If we take a negative entropy again, the second term on the right hand side will become positive. Hence the total value of

$\Delta G$ will be positive. But Gibb's free energy for a spontaneous reaction can't be positive. Hence statement 1 is not correct.

Statement 2 is correct. Gibb's free energy for a spontaneous reaction is negative.

Option D is the correct answer.

How much energy is needed to melt

$100g$ of water?

$\Delta {H}_{v}=540\ cal/g$

$\Delta {H}_{f}=80\ cal/g$

${C}_{water}=1\ cal/{g}^{o}C$

0%
$640cal$
0%
$8000cal$
0%
$54,000cal$
0%
$180cal$

Explanation

Latent heat of fusion

$=80cal/g$

To melt 100gm ,we have

$(80)(100)=8000$ calories of heat.

Which of the following conditions guarantee a spontaneous reaction?

0%
Positive $\triangle H$ , positive $\triangle S$
0%
Positive $\triangle H$ , negative $\triangle S$
0%
Negative $\triangle H$ , negative $\triangle S$
0%
Negative $\triangle H$ , positive $\triangle S$
0%
None of the above

Explanation

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.

$\Delta S$ 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

At what point of the heating curve are intermolecular forces at their greatest level?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Given that

$\Delta G = \Delta H - T \Delta S$ , how is the spontaneity of an endothermic reaction expected to change with decreasing

$T$ ?

0%
Becomes less spontaneous
0%
Becomes more spontaneous
0%
Does not change
0%
Decreases at first but then increases
0%
Insufficient information to make a conclusion

Explanation

For an endothermic reaction,

$\Delta H$ is positive. And for spontaneity

$\Delta G$ should be negative. Now, if temperature

$T$ is decreased but it has a positive value, whether

$\Delta G$ should be negative or not will depend on the value of

$\Delta S$ (whether

$\Delta S$ is positive or not).

If temperature

$T$ is decreased and has a negative value, then again, whether

$\Delta G$ should be negative or not will depend on the value of

$\Delta S$ (whether

$\Delta S$ is positive or not).

Because in both the cases

$\Delta H$ is positive, so whether

$\Delta G$ should be negative or not will depend on the value of

$\Delta S$ .

Thus, option E is the correct answer. The information is insufficient to make a conclusion.

Which selection correctly shows units for specific heat capacity?

0%
$Cal \cdot g\cdot ^{o}C$
0%
$Cal \cdot g/ ^{o}C$
0%
$J/(g \cdot ^{o}C)$
0%
$g/J \cdot ^{o}C$

Explanation

$Q=nC\Delta T$

Therefore,

$C=\cfrac {Q}{n\Delta T}$

$\Rightarrow \cfrac {J}{g\times ^oC}$

What is the specific heat of a

$2.0\ g$ sample of metal which requires the addition of

$8\ J$ to be heated from

$293\ K$ to

$303\ K$ ?

0%
$0.4$
0%
$0.02$
0%
$0.6$
0%
$2.5$

Explanation

According to the formula,

$Q=mC\triangle T$

$Q=8J=$ amount of heat absorbed or released

$m=2.0g=mass$

$\triangle T$ =change in temperature

$=(303-293)K=10$

Therefore,

$8J=2.0g\times C\times 10K$

$C=\cfrac {8}{20.0}=\cfrac {0.8}{2}=0.4$

$Jg^{-1}K^{-1}$

Hence, option

$A$ is correct.

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

0%
Voltage would increase since more ions can give up electrons.
0%
Voltage will decrease since there will be less $Cu^{2+}$ ions in solution.
0%
Voltage would stay the same since $NaOH$ is a strong electrolyte and will exist as a spectator ion.
0%
Voltage would increase and then decrease as the addition of new ions reach equilibrium.

Explanation

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

To which portion of the heating curve for water does

$\Delta {H}_{fus}$ apply?

0%
$A-B$
0%
$B-C$
0%
$C-D$
0%
$D-E$

Explanation

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.

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?

0%
Dissolution of sodium hydroxide.
0%
Creating a dilution of hydrochloric acid.
0%
The burning of $C_{20}H_{42}$ .
0%
The synthesis of silver oxide.

Explanation

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

Which portion of the heating curve for water shown above would there be both liquid and solid water present?

0%
$A-B$
0%
$B-C$
0%
$C-D$
0%
$D-E$

To which portion of the heating curve for water does

$C=1.00 \ cal/{g}^{o}C$ apply?

0%
$A-B$
0%
$B-E$
0%
$C-D$
0%
$A-F$

The formation of carbon dioxide by reacting Carbon with excess Oxygen, is exothermic. What will be the sign of

$\Delta H$ ?

0%
$\Delta H$ is positive
0%
$\Delta H$ is negative
0%
$\Delta H$ is zero
0%
None of these

Explanation

$C$

$+$

$O_2$

$=$

$CO_2$

$+$

$Heat$
When carbon dioxide is produced by reacting Carbon and Oxygen,

$-393kJ$ of heat is produced or evolved. Thus, the reaction is exothermic in nature as heat is released. Hence,

$\Delta H$ will be negative.

How will the temperature of

$10.0\ g$ of iron change if

$45\ J$ is added to the metal?
Iron has a specific heat of

$0.45\ J/(g^oC)$ .

0%
$+20^{o}C$
0%
$+10^{o}C$
0%
$+100^{o}C$
0%
$+1^{o}C$

Explanation

$Q=nC\triangle T$

$Q=45J$ ;

$n=10.0g$ ;

$C=0.45J/g^oC$

Therefore,

$45=nC\triangle T$

$\Rightarrow \cfrac {45}{10.0\times 0.45}=\triangle T$

$\Rightarrow 10=\triangle T$

To which portion of the heating curve for water does

$\Delta {H}_{vap}$ apply?

0%
$A-B$
0%
$B-C$
0%
$C-D$
0%
$D-E$

Explanation

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.

$C_v$ for helium gas

$(He)$ is

$(in J \ mol^{-1} \ K^{-1})$

0%
2.5
0%
10.5
0%
1.5
0%
12.5

Explanation

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

$dU=C_{V}dT$

$C_{V}=dU/dT$

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

$C_{V}=\frac{3\times8.31}{2}=12.5Jmol^{-1}K^{-1}$

Carbon reacts with oxygen to produce carbon dioxide which is an exothermic reaction, where

$394\ kJ/mol$ of heat is produced. What will be the value of

$\Delta H$ when

$24$ grams of carbon reacts?

0%
$+394\ kJ$
0%
$-394\ kJ$
0%
$+788\ kJ$
0%
$-788\ kJ$

Explanation

$C$

$+$

$O_2$

$=$

$CO_2$

$+$

$Heat$ .

Thus, we see that

$1$ mole or

$12$ grams of carbon gives out

$394\ kJ$ of heat.

$1\ g$ of carbon gives out

$\dfrac{394}{12}$

$kJ$ of heat.

$24\ g$ of carbon will give out

$\dfrac{394}{12}$

$\times$

$24$ , that is

$788\ kJ$ of heat.

Since the reaction is exothermic in nature, that is heat will be given out or produced.

Hence, the value of

$\Delta H$ will be negative. Thus,

$\Delta H$ is

$-788\ kJ$ .

The formation of nitric oxide from Nitrogen and Oxygen is endothermic in nature. What will be the sign of

$\Delta H$ ?

0%
$\Delta H$ is positive
0%
$\Delta H$ is negative
0%
$\Delta H$ is zero
0%
None of these

Explanation

$N_2$

$+$

$O_2$

$+$

$Heat$

$=$

$2NO$
Almost

$181kJ$ of heat is absorbed during the reaction of Nitrogen and Oxygen to produce nitric oxide. Since heat is absorbed, the reaction is endothermic in nature. So, the value of

$\Delta H$ will be positive.

A balanced stoichiometric chemical equation which includes a change in enthalpy, that is

$\Delta H$ , is known as:

0%
precipitation reaction
0%
acid-base neutralization
0%
thermochemical reaction
0%
oxidation-reduction reaction

A piece of ice kept at room temperature melts of its own. This reaction is governed by which law?

0%
Second law of thermodynamics
0%
First law of thermodynamics
0%
Third law of thermodynamics
0%
Zeroth law of thermodynamics

Carbon reacts with Sulphur to give carbon disulphide, which is an endothermic reaction, where

$92kJ/mol$ of heat is absorbed. What will be the value of

$\Delta H$ , when

$24$ grams of carbon reacts?

0%
$-92kJ$
0%
$+92kJ$
0%
$-184kJ$
0%
$+184kJ$

Explanation

The reaction can be written as,

$C$

$+$

$2S$

$+$

$Heat$

$\rightarrow$

$CS_2$ .
Thus, we see that

$1mol$ or

$12\ grams$ of Carbon absorbs

$92kJ$ of heat.

$1\ gram$ of Carbon will absorb

$\dfrac{92}{12}$

$kJ$ of heat
Hence,

$24\ grams$ of Carbon will absorb

$\dfrac{92}{12}$

$*$

$24$ , that is

$184kJ$ of heat.
Since, heat is absorbed, the reaction is endothermic in nature, and the value of

$\Delta H$ will be positive, that is

$+184kJ$ .

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

0%
Adiabatic
0%
Isochoric
0%
Isobaric
0%
Isothermal

Explanation

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.

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?

0%
$-786\space kJ/mol$
0%
$788\space kJ/mol$
0%
$388\space kJ/mol$
0%
$-508\space kJ/mol$

Explanation

$\Delta H_{f^0} = \Delta H_{sub} + IE+\Delta H_{diss} + EA + U$

$\Delta H_{f^0} = 108+496+122-349-788 = 411\, kJ/mol$

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.

For a thermodynamics process to be reversible, the temperature difference between hot body and the working substance should be

0%
zero
0%
minimunm
0%
maximum
0%
infinity

Thermochemical equation does not indicate the amount of heat evolved or absorbed.

0%
True
0%
False

Explanation

It is sometimes convenient to provide the value for $\Delta { H }_{ rxn }$ along with the balanced chemical equation for a reaction (also known as a thermochemical equation):

$2{ H }_{ 2(g) }+{ O }_{ 2(g) }\longrightarrow 2{ H }_{ 2 }{ O }_{ (g) },\Delta { H }_{ rxn }=-483.6kJ$

In a thermodynamic process, a system absorbs 2 k cal of heat and at the same time does 800 J of work. The change in internal energy of the system is

0%
7600 J
0%
2000 J
0%
4800 J
0%
-7600 J

Explanation

In a thermodynamic process, a system absorbs 2 k cal of heat and at the same time does 800 J of work

By calorie to joules conversion

we have

$1 cal =4.2J$

$\therefore$

$2Kcal=8400 J$

Hence by first law of thermodynamics

we have,

$\delta U=\delta H -W=8400-800=7600J$

Consider the following processes

$\triangle H(kJ/mol)$

$\dfrac {1}{2}A\rightarrow B + 150$

$3B \Rightarrow 2C + D - 125$

$E + A \rightarrow 2D + 350$

For

$B + D \rightarrow E + 2C, \triangle H$ will be:

0%
$+525\ kJ/mol$
0%
$-175\ kJ/mol$
0%
$-325\ kJ/mol$
0%
$+325\ kJ/mol$

Explanation

$\dfrac {1}{2}A\rightarrow B; \triangle H = 150\ kJ/mol ... (i)$

$3B \rightarrow 2C + D; \triangle H = + 350 \ kJ/mol .... (ii)$

$E + A \rightarrow 2D; \triangle H = + 350\ kJ/mol ... (iii)$

$[2\times (i) + (ii)] - (iii)$ , we get

$B + D \rightarrow E + 2C$

$\therefore \triangle H = 150\times 2 + (-125) - 350$

$= -175\ kJ/mol$

Which of the properties are suitable for a cooking utensil?

0%
High specific heat
0%
Low specific heat
0%
High conductivity
0%
Low conductivity

Consider the following changes

$M\left( s \right) \longrightarrow M\left( g \right)$ ...(1)

$M\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +2{ e }^{ - }$ ....(2)

$M\left( g \right) \longrightarrow { M }^{ + }\left( g \right) +{ e }^{ - }$ ....(3)

${ M }^{ + }\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +{ e }^{ - }$ ...(4)

$M\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +2{ e }^{ - }$ ...(5)
The second ionisation energy of

$M$ could be determined from the energy values associated with:

0%
$1+2+4$
0%
$1+5-3$
0%
$2+3-4$
0%
$5-3$

Explanation

The amount of energy required to take out an electron from the monopositive cation is called second ionisation energy

$M\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +2{ e }^{ - }$ (v)

$M\left( g \right) \longrightarrow { M }^{ + }\left( g \right) +{ e }^{ - }$ (iii)
On subtracting equation (iii) from equation (v) we get,

$M\longrightarrow { M }^{ 2+ }+{ e }^{ - }$ .

Thermodynamic condition for irreversible spontaneous process at constant '

$T$ ' and '

$P$ ' is:

0%
$\Delta G< 0$
0%
$\Delta G=0$
0%
$\Delta G> 0$
0%
Both (B) and (C)

Explanation

For spontaneous process

$\Delta G<0$ and for non spontaneous process

$\Delta G>0$ .

Option A is correct.

What is the thermodynamic condition for the feasibility of a process?

0%
$(\Delta S)_{sys} > 0$
0%
$(\Delta G)_{sys} > 0$
0%
$(\Delta S)_{sys} + (\Delta S)_{surr} > 0$
0%
$(\Delta G)_{sys} + (\Delta G)_{surr} < 0$

Specific heat of a substance can be :

0%
zero
0%
positive
0%
infinity
0%
negative

Explanation

Specific heat is the amount of heat that is required to raise the temperature of 1 gram of a substance by 1 degree Celcius.

Specific heat

$C=\cfrac { \partial Q }{ \triangle T\triangle m } { kJ }/{ kgK }$

For isothermal process,

$\triangle T=0$

Hence

$C=\infty$

For isothermal process,

$\partial Q=0$

Hence

$C=0$

When heat is absorbed by the system, q is positive. So specific heat will be positive.

When heat is absorbed from the system, q is negative. So specific heat will be negative.

For the reaction

$N_2+3X_2 \rightarrow 2NX_3$ where

$X = F, Cl$ (the average bond energies are

$F-F = 155 \ kJ \ mol^{-1}$ ,

$N-F = 272 \ kJ \ mol^{-1}$ ,

$Cl-Cl=242\ kJ \ mol^{-1}$ ,

$N-Cl=200\ kJ \ mol^{-1}$ and

$N \equiv N = 941 kJ\, mol^{-1}$ ). The heats of formation of

$NF_3$ and

$NCl_3$ in

$kJ\ mol^{-1}$ , respectively, are closest to:

0%
$- 226$ and $+ 467$
0%
$+ 226$ and $-467$
0%
$-151$ and $+ 311$
0%
$+ 151$ and $-311$

Explanation

$N_2 + 3X_2 \rightarrow 2NX_3$

Energy to break 1 mole of

$N_2$ bond

$= 941 \ kJ mol^{-1}$

Energy to break 3 mole of

$F_2$ bond

$= 3 \times 155 \ kJ \ mol^{-1}$

Energy released caused of bond formation of 2 moles of 3 bonds

$N-F = 2 \times 3 \times 272$

Heat of formation of

$NF_3$ by Hess law

$= 941+3 \times 155 - (2 \times 3 \times 272) = -226 \ kJ \ mol^{-1}$

Similarly for

$NCl_3$ ,

Just replacing the necessary values in previous calculations, heat of formation of

$NCl_3 = 941 + 3 \times 242 - (2 \times 3 \times 200) = +467 \ kJ mol^{-1}$ .

Hence, option

$A$ is correct.

Which of the following processes is always non - feasible?

0%
$\Delta H > 0, \Delta S > 0$
0%
$\Delta H < 0, \Delta S > 0$
0%
$\Delta H > 0, \Delta S < 0$
0%
$\Delta H < 0, \Delta S < 0$

Explanation

In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum or reversible work that may be performed by a thermodynamic system at a constant temperature and pressure (isothermal, isobaric).

$\Delta G=\Delta H-T\Delta S$

A reaction is non-feasible if $\Delta G$ is positive. So, if $\Delta H$ is positive and $\Delta S$ is negative then $\Delta G$ is always positive irrespective of their magnitude.

For a chemical reaction,

$\Delta S = - 0.035\ kJ/K$ and

$\Delta H = - 20k J$ . At what temperature does the reaction turn non-spontaneous?

0%
$<5.14\ K$
0%
$<57.14\ K$
0%
$<571.4\ K$
0%
$<5714.0\ K$

Explanation

For a reaction to turn non-spontaneous:

$\Delta G > 0$

$\Delta H - T\Delta S >0$

$\dfrac{\Delta H}{\Delta S} > T$

$\dfrac{20000}{35} > T$

$T < 571.4\ K$

Consider the Born-Haber cycle for the formation of an ionic compound given below and identify the compound

$\left(Z\right)$ formed.

0%
${M}^{+}{X}^{-}$
0%
${M}^{-}{X}_{\left(s\right)}^{-}$
0%
$MX$
0%
${M}^{+}{X}_{\left(g\right)}^{-}$

Explanation

Based on the above equations it is clear that the cation formed is

$M^{+}$ and anion formed is

$X^{-}$ . So, the compound formed is

$M^{+}X^{-}$ .

The criterion for a spontaneous process is :

0%
$\triangle G > 0$
0%
$\triangle G < 0$
0%
$\triangle G = 0$
0%
$\triangle S_{total} < 0$

Explanation

For a spontaneous process to occur, the free energy should always decrease from its initial value.

So,

$\Delta G<0$ .

If enthalpies of formation for

$C_2H_4(g)$ ,

$CO_2(g)$ and

$H_2O(l)$ at

$25^o$ C and

$1$ atm pressure be

$52$ ,

$-394$ and

$-286$

$kJ \ mol^{-1}$ respectively, then the enthalpy of combustion of

$C_2H_4(g)$ will be:

0%
$-141.2$ kJ $mol^{-1}$
0%
$-1412$ kJ $mol^{-1}$
0%
$+141.2$ kJ $mol^{-1}$
0%
$+1412$ kJ $mol^{-1}$

Explanation

Given,

$2C(g)+2H_2(g)\rightarrow C_2H_4(g); \Delta H_1=52kJmol^{-1}$ .......(i)

$C(g)+O_2(g)\rightarrow CO_2(g); \Delta H_2=-394kJmol^{-1}$ .......(ii)

$H_2(g)+\dfrac{1}{2}O_2(g)\rightarrow H_2O(g); \Delta H_3-286kJmol^{-1}$ ..........(iii)

Reaction involved for combustion of

$C_2H_4(g)$ is,

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O; \Delta H=?$ .....(iv)

$2\times[Eq. (ii)+Eq. (iii)]-Eq.(i),$ we get Eq.(iv)

$\therefore \Delta =2[\Delta H_2+\Delta H_3]-\Delta H_1$

$=2[-394-286]-52$

$=-1360 -52=-1412$

$kJ\ mol^{-1}$ .

Hence,option B is correct.

What is the enthalpy of the disproportion of MgCl if the enthalpy of formation of hypothetical MgCl is -125 kJ/mol and the

$MgCl_2$ is -642 kJ mol

$^{ -1 }$ ?

0%
-767 kJ mol $^{ -1 }$ .
0%
767 kJ mol $^{ -1 }$ .
0%
-392 kJ mol $^{ -1 }$ .
0%
392 kJ mol $^{ -1 }$ .

Explanation

$Mg(s) + Cl_2 (g) \longrightarrow MgCl_2(s)$

$\triangle H_1$ = - 642 kJ mol

$^{ -1 }$ .

$Mg(s) + \frac {1}{2} Cl_2 (g) \longrightarrow MgCl (s)$

$\triangle H_1$ = - 125 kJ mol

$^{ -1 }$ .

$2MgCl \longrightarrow MgCl_2 + Mg$

$\Rightarrow \Delta H = \Delta H_1 - 2 \Delta H_2$

= -642 - 2

$\times$ (-125) = -392 kJ mol

$^{ -1 }$ .

Hence, the correct option is

$C$

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 4 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 4 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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