MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 5

In biological systems flow of energy occurs through :

0%
Loss of electrons.
0%
Gain of electrons.
0%
Both A and B
0%
Photons

Explanation

$\text{Option B is correct.}$

$\text{Gaining of electrons means releasing of energy and becoming more stable for a substance or element. }$

$\text{While it needs enegy for lossing of electrons.}$

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

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

$\Delta { U }^{ o }$ of combustion of methane is

$-X$

$kJ$

${ mol }^{ -1 }$ . The value of

$\Delta { H }^{ o }$ is:

0%
$=\Delta { U }^{ o }$
0%
$>\Delta { U }^{ o }$
0%
$<\Delta { U }^{ o }$
0%
$=0$

Explanation

Reaction;

$C{ H }_{ 4 }\left( g \right) +2{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right)$

Now,

$\Delta { n }_{ g }=\left( { n }_{ p }-{ n }_{ r } \right) =1-3=-2$

Again,

$\Delta { H }^{ o }=\Delta { U }^{ o }+\Delta { n }_{ g }RT$

$\Delta { H }^{ o }=-X-2RT$

$\therefore \Delta { H }^{ o }<\Delta { U }^{ o }$

The molar enthalpy change for

$H_2O(1)\rightleftharpoons H2O(g)$ at

$373$ K and

$1$ atm is

$41$ kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of

$1$ mol of water at

$373$ K and

$1$ atm in kJ

$mol^{-1}$ is

$?$

0%
$30.2$
0%
$41.0$
0%
$48.1$
0%
$37.9$

Explanation

$W=-nRT=1(8.314\times 373)\times 10^{-3}=-3.10 KJ$

$q=41KJ$

$\Delta U=37.9KJ$

A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If

$\Delta$ H is the enthalpy change and

$\Delta$ E is the change in internal energy, then:

0%
$\Delta H > \Delta E$ .
0%
$\Delta H < \Delta E$ .
0%
$\Delta H = \Delta E$ .
0%
the relationship depends on the capacity of the vesssel.

Explanation

$\displaystyle 2CO +O_2\rightarrow 2CO_2$

$\displaystyle \Delta n$ = no. of moles of gaseous product

$-$ no. of moles of gaseous reactant

$\displaystyle \Delta n = 2 - (2+1) = -1$

$\displaystyle \Delta H = \Delta E + \Delta nRT$

$\displaystyle \Delta H = \Delta E + (-1)RT$

$\displaystyle \Delta H = \Delta E - RT$

$\displaystyle \Delta H < \Delta E$

Hence,option B is correct.

"The change of enthalpy of a chemical reaction is the same whether the reaction takes place in one step or in several steps". This law was presented by:

0%
Hess
0%
La Chatelier
0%
Kirchhoff
0%
Lavoisier and Laplace

The enthalpies of combustion of

${C}_{(graphite)}$ and

${C}_{(diamond)}$ are

$-393.5$ and

$-395.4kJ/mol$ respectively. The enthalpy of conversion of

${C}_{(graphite)}$ to

${C}_{(diamond)}$ in

$kJ/mol$ is:

0%
$-1.9$
0%
$-788.9$
0%
$1.9$
0%
$788.9$

Explanation

The combustion reactions for

$C_{(graphite)}$ and

$C_{(diamond)}$ can be written as,

$C_{(graphite)} + O_2 \rightarrow CO_2$

$\Delta H_1 = -393.5 \dfrac{KJ}{mol}$ (1)

$C_{(diamond)} + O_2 \rightarrow CO_2$

$\Delta H_2 = -395.4 \dfrac{KJ}{mol}$ (2)
On subtracting reaction (2) from reaction (1) we will get,

$C_{(graphite)} \rightarrow C_{(diamond)}$ with

$\Delta H = \Delta H_1 - \Delta H_2 = -393.5 - (-395.4) = 1.9\dfrac{KJ}{mol}$
Hence, answer is option C.

For the reactions,
(i)

${H}_{2}(g)+{Cl}_{2}(g)=2HCl(g)+x$

$kJ$
(ii)

$2HCl(g)={H}_{2}(g)+{Cl}_{2}(g)-y$

$kJ$
which one of the following statements is correct?

0%
$x-y> 0$
0%
$x-y< 0$
0%
$x-y=0$
0%
None of these

Explanation

Since 1

$^{st}$ reaction is opposite of the 2

$^{nd}$ reaction.
So, on adding both reactions we will get,

$x - y = 0$
Hence, answer is option

$C$ .

From the thermochemical reactions,

${C}_{(graphite)}+\cfrac{1}{2}{O}_{2}\rightarrow CO;\Delta H=-110.5kJ$

$CO+\cfrac{1}{2}{O}_{2}\rightarrow {CO}_{2};\Delta H=-283.2kJ$
the heat of reaction of

${C}_{(graphite)}+{O}_{2}\rightarrow {CO}_{2}$ is:

0%
$+393.7kJ$
0%
$-393.7kJ$
0%
$-172.7kJ$
0%
$+172.7kJ$

Explanation

Given reactions are,

$C_{(graphite)} +\dfrac{1}{2}O_2 \rightarrow CO$

$\Delta H = -110.5KJ$

$CO + \dfrac{1}{2}O_2 \rightarrow CO_2$

$\Delta H = -283.2KJ$
Since on adding both the above reaction we will get a final reaction

$C_{(graphite)} +O_2 \rightarrow CO_2$
which is the same equation whose heat of reaction is being asked.
Since, on adding two reactions the heat of reactions gets added.
Hence, the heat of reaction for the final reaction will be,

$\Delta H = \Delta H_1 + \Delta H_2 = -110.5 - 283.2 = 393.7KJ$
Hence, answer is option B.

Spontaneous reactions are :

0%
Endergonic
0%
Exergonic
0%
Energy neutral
0%
Exer-endergonic reactions

Explanation

Exergonic reactions are the reactions in which energy is released. Initially, exergonic reaction requires activation energy to star the reaction process. Once this activation energy is reached, the reaction proceeds to break bonds and form new bonds. Energy is released as the reaction proceeds. This led to net gain of energy in the surrounding system and a net loss in energy from the reaction system. Hence

$\triangle G$ is negative and reaction is spontaneous.

Hence we can say that spontaneous reactions are exergonic.

For the reactions,

(i)

${H}_{2}(g)+{Cl}_{2}(g)=2HCl(g)+x$

$kJ$

(ii)

${H}_{2}(g)+{Cl}_{2}(g)=2HCl(l)+y$

$kJ$

which one of the following statements is correct?

0%
$x> y$
0%
$x< y$
0%
$x-y=0$
0%
$x=y$

Explanation

(i)

${H}_{2}(g)+{Cl}_{2}(g)=2HCl(g)+x$

$kJ$

(ii)

${H}_{2}(g)+{Cl}_{2}(g)=2HCl(l)+y$

$kJ$

In 2

$^{nd}$ reaction the product

$HCl$ formed is in liquid state while in 1

$^{st}$ reaction it is in gaseous state, so their will be some energy used in convert

$HCl$ liquid into

$HCl$ gas due to which the heat evolved in the reaction is less.

Hence,

$x<y$ . the answer is option

$B$ .

$\Delta {H}_{f}^{o}$ for

${H}_{2}{O}_{2}(l)$ and

${H}_{2}O(l)$ are

$-188kJ$

${mol}^{-1}$ and

$-286kJ$

${mol}^{-1}$ , what will be the enthalpy change of the reaction?

$2{H}_{2}{O}_{2}(l)\rightarrow 2{H}_{2}O(l)+{O}_{2}(g)$

0%
$146kJ$ ${mol}^{-1}$
0%
$-196kJ$ ${mol}^{-1}$
0%
$-494kJ$ ${mol}^{-1}$
0%
$-98kJ$ ${mol}^{-1}$

Explanation

For the given reaction,

$2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)$
enthalpy change of the reaction =

$\Delta H$ =

$2\times \Delta H_{f(H_2O)} - 2\times \Delta H_{f(H_2O_2)}$ (since,

$\Delta H_{f(O_2)} = 0$ )
hence,

$\Delta H = 2\times (-286) -2\times (-188) = -196KJ\text{ mol}^{-1}$
Hence, answer is option

$B$ .

The heats of combustion of yellow phosphorus and red phosphorus are

$-9.91kJ$ and

$-8.78kJ$ respectively. The heat of transition of yellow phosphorus to red phosphorus is:

0%
$-18.69kJ$
0%
$+1.13kJ$
0%
$+18.69kJ$
0%
$-1.13kJ$

Explanation

Since the conversion reaction can be written as

$P_{(Yellow)} \rightarrow P_{(Red)}$

Hence, the heat of reaction of above reaction (

$\Delta H$ ) can be written as,

$\Delta H = \text{heat of combustion of } P_{(Yellow)} - \text{heat of combustion of } P_{(Red)} = -9.91KJ - (-8.78KJ) = -1.13KJ.$

Hence, the answer is option D.

Given:

$C(s)+{O}_{2}(g)\rightarrow {CO}_{2}(g)+94.2kcal$

${H}_{2}(g)+\cfrac{1}{2}{O}_{2}(g)\rightarrow {H}_{2}O(l)+68.3kcal$

${CH}_{4}+2{O}_{2}(g)\rightarrow {CO}_{2}(g)+2{H}_{2}O(l)+210.8kcal$
The heat of formation of methane in

$kcal$ will be:

0%
$45.9$
0%
$47.8$
0%
$20.0$
0%
$47.3$

Explanation

First reaction is formation reaction of carbon dioxide. Hence,

$\text{heat of formation of carbon dioxide = 94.2Kcal.}$
Second reaction is formation reaction of water. Hence,

$\text{heat of formation of water = 68.3kCal.}$
Since,

$\Delta H$ for third reaction =

$210.8Kcal$
Also,

$\Delta H$ =

$\Delta H_{f(CO_2)} + 2\times \Delta H_{f(H_2O)} - \Delta H_{f(CH_4)}$

$210.8 = 94.2 + 2\times 68.3 - \Delta H_{f(CH_4)}$
=>

$\Delta H_{f(CH_4)} = 94.2 +136.6 -210.8 = 20Kcal.$
Hence, answer is option

$C$ .

The heats of combustion of rhombic and monoclinic sulphur are

$-70960$ and

$-71030$ calorie respectively. What will be the heat of conversion of rhombic sulphur to monoclinic sulphur?

0%
$-70960\ cal$
0%
$-71030\ cal$
0%
$+70\ cal$
0%
$-70\ cal$

Explanation

The conversion reaction of rhombic sulphur to monoclinic sulphur is given by,

$S_{rhombic} \rightarrow S_{monoclinic}$

Hence,

$\Delta H_{conversion} = \Delta H_{\text{combustion of S(rhombic)}} -\Delta H_{\text{combustion of S(monoclinic)}}$

$\Delta H_{conversion} =[( -70960) - (-71030)] = 70cal.$

$\therefore$ answer is option

$C$ .

For an exothermic reaction to be spontaneous:

0%
temperature must be high
0%
temperature must be zero
0%
temperature may have any magnitude
0%
temperature must be low

Explanation

Gibb's free energy equation :

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

For spontaneous reaction :

$\Delta G<0$

For exothermic reaction :

$\Delta H<0$

$\Delta H<0$ &

$\Delta S>0$ , the temperature can have any value for reaction to be spontaneous. Because,

$\Delta H<0\quad \& \quad T\Delta S>0$ , So the

$\Delta G<0$ .

For which reaction from the following,

$\Delta S$ will be maximum?

0%
$Ca(s)+\cfrac{1}{2}{O}_{2}(g)\rightarrow CaO(s)$
0%
$Ca{CO}_{3}(s)\rightarrow CaO(s)+{CO}_{2}(g)$
0%
$C(s)+{O}_{2}(g)\rightarrow {CO}_{2}(g)$
0%
${N}_{2}(g)+{O}_{2}(g)\rightarrow 2NO(g)$

Explanation

Explanation:

The entropy change depends upon the number of gaseous molecules.

For entropy change to be maximum, the difference between the moles of gaseous products and gasous reactants should be maximum.

$\Delta {n_{g}}$

$=Sum\ of\ the\ number\ of\ moles\ of\ gaseous\ products - Sum\ of\ the\ number\ of\ moles\ of\ gaseous\ reactants$

$\Delta {n_g}$ for the given reactions are given below,

$\Delta {n_g}=0.5$

$\Delta {n_g}=1$

$\Delta {n_g}=0$

$\Delta {n_g}=0$ .

Final Answer: Since

$\Delta {n_g}$ value is the maximum for reaction

$B$ , hence option

$B$ is the correct answer.

The value of

$\Delta H$ and

$\Delta S$ for a reaction are respectively

$30kJ$

${mol}^{-1}$ and

$100J{K}^{-1}$

${mol}^{-1}$ . Then temperature above which the reaction will become spontaneous is:

0%
$300K$
0%
$30K$
0%
$100K$
0%
${300}^{o}C$

Explanation

Gibb's free energy equation :

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

For reaction to be spontaneous,

$\Delta G<0$

$\therefore \quad \Delta H-T\Delta S<0$

$\therefore \quad \left( 30\times { 10 }^{ 3 } \right) -T\times \left( 100 \right) <0$

$\therefore \quad 30000<100T$

$\therefore \quad 300<T$

$\therefore \quad T>300K$

Therefore, for reaction to be spontaneous, temperature should be above 300 K.

Which of the following thermodynamics relation is correct?

0%
$dG=VdP-SdT$
0%
$dU=PdV+TdS$
0%
$dH=-Vdp+TdS$
0%
$dG=VdP+SdT$

Explanation

$dG=dH-TdS-SdT$ ...

$(G=H-TS)$

$dH=dU+PdV+VdP$ ...

$(H=U+PV)$
and

$dU=TdS-PdV$

$\therefore$

$dG=(TdS-PdV)+PdV+VdP-TdS-SdT$

$dG=VdP-SdT$

On combustion carbon forms two oxides

$CO$ and

${CO}_{2}$ , heat of formation of

${CO}_{2}$ is

$-94.3\ kcal$ and that of

$CO$ is

$-26.0 \ kcal$ . Heat of combustion of carbon is:

0%
$-26.0\ kcal$
0%
$-68.3\ kcal$
0%
$-94.3\ kcal$
0%
$-120.3\ kcal$

The heat of combustion of ethanol determined in a bomb calorimeter is

$-670.48kcal$

${mol}^{-1}$ at

$298K$ . What is the

$\Delta U$ at

$298K$ for the reaction?

0%
$-760\ kcal$ ${mol}^{-1}$
0%
$-670.48\ kcal$ ${mol}^{-1}$
0%
$+760\ kcal$ ${mol}^{-1}$
0%
$+670.48\ kcal$ ${mol}^{-1}$

Explanation

Bomb calorimeter works on a basic principle that involves combustion of the organic compound at constant volume constraint.
Since, at constant volume the heat of combustion of compound = change in internal energy of that compound (as pressure-volume work will be zero).
Hence,

$\Delta U$

$= \text{heat of combustion in a bomb calorimeter} = -670.48\text{ Kcal mol}^{-1}.$
Hence, answer is option B.

$\Delta { G }^{ o }$ for the reaction

$x+y\rightleftharpoons z$ is

$-4.606kcal$ . The value of equilibrium constant of the reaction at

${227}^{o}C$ is:
(

$R=2.0cal.{mol}^{-1}{K}^{-1}$ )

0%
$100$
0%
$10$
0%
$2$
0%
$1$

In the conversion of limestone to lime,

$Ca{CO}_{3}(s)\rightarrow CaO(s)+{CO}_{2}(g)$
the values of

$\Delta {H}^{o}$ and

$\Delta {S}^{o}$ are

$+179.1kJ{ mol }^{ -1 }$ and

$160.2kJ{ mol }^{ -1 }$ respectively at

$298K$ and

$1$ bar. Assuming that,

$\Delta {H}^{o}$ and

$\Delta {S}^{o}$ do not change with temperature; temperature above which conversion of limestone to lime will be spontaneous is:

0%
$1118K$
0%
$1008K$
0%
$1200K$
0%
$845K$

Explanation

${ CaCO }_{ 3 }\left( s \right) \rightarrow CaO\left( s \right) +{ CO }_{ 2 }\left( g \right)$

${ \Delta H }^{ 0 }=+179.1KJ/mol$

${ \Delta S }^{ 0 }=160.2KJ/mol$

${ \Delta G }^{ 0 }={ \Delta H }^{ 0 }-T\Delta { S }^{ 0 }$

For spontaneous process,

${ \Delta G }^{ 0 }<0$

$\therefore \quad { \Delta H }^{ 0 }-T{ \Delta S }^{ 0 }<0$

$\therefore \quad { 10 }^{ 3 }\times 179.1<160.2\times T$

$\therefore \quad T>\dfrac { 179.1\times { 10 }^{ 3 } }{ 160.2 }$

$\therefore \quad \boxed { T>1118K }$

Therefore, above

$1118K$ , the reaction will be spontaneous.

Considering entropy(s) as a thermodynamic parameter, the criterion for the spontaneity of any process is:

0%
$\Delta {S}_{system}+\Delta {S}_{surroundings}> 0$
0%
$\Delta {S}_{system}-\Delta {S}_{surroundings}> 0$
0%
$\Delta {S}_{system}> 0$ only
0%
$\Delta {S}_{surroundings}> 0$ only

Explanation

Entropy (S) is the measure of randomness or disorder of the molecules. It is an extensive property and a state function.

A process is spontaneous if and only if the entropy of the universe increases.

Therefore, for spontaneous process,

${ \Delta S }_{ universe }>0\quad or\quad { \Delta S }_{ system }+{ \Delta S }_{ surroundings }>0$

For the process,

${H}_{2}O(l) (1bar,373K)\rightarrow {H}_{2}O(g)(1bar;373K)$
The correct set of thermodynamics parameters is:

0%
$\Delta G=0,\Delta S=+ve$
0%
$\Delta G=0,\Delta S=-ve$
0%
$\Delta G=+ve,\Delta S=0$
0%
$\Delta G=-ve,\Delta S=+ve$

Explanation

${ H }_{ 2 }O\left( l \right) \rightleftharpoons { H }_{ 2 }O\left( g \right)$

It is an equilibrium process.

Therefore,

$\Delta G=0$

Now,

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

where

$\Delta H$ is the enthalpy of vaporization &

$\Delta H>0$

Therefore, for

$\Delta G=0$ , the value of

$\Delta S$ should be positive. Also, the

$\Delta S$ is always when a molecule is energized. Here, the liquid

${ H }_{ 2 }O$ is energized to gaseous

${ H }_{ 2 }O$ . So,

$\Delta S$ is +ve.

Calculate

$\triangle G^{\circ}$ for the following cell reaction:

$Zn(s) + Ag_{2}O(s) + H_{2}O(l) \rightleftharpoons Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq)$

$E^{\circ}_{Ag^{+}/Ag} = + 0.80V$ and

$E^{\circ}_{Zn^{2+}/ ZN} = -0.76V$ .

0%
$-305\ kJ/mol$
0%
$-301\ kJ/mol$
0%
$+305\ kJ/mol$
0%
$+301\ kJ/mol$

Hess law is based on:

0%
Law of conservation of mass
0%
Law of conservation of energy
0%
Enthalpy is a state function
0%
None of these

Which of the following is true for spontaneous process?

0%
$\Delta G> 0$
0%
$\Delta G< 0$
0%
$\Delta G= 0$
0%
$\Delta G=T\Delta S$

Explanation

The process which proceeds in a particular direction by its own under a given set of conditions without outside help is called sponataneous process. All natural processes are spontaneous.

Gibb's free energy,

$\Delta G<0\quad \Rightarrow$ for spontaneous process

$\Delta G>0\quad \Rightarrow$ for non-spontaneous process

$\Delta G=\quad \Rightarrow$ for equilibrium process

The Haber's process for production of ammonia involves the equilibrium:

${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons 2{NH}_{3}(g)$
Assuming

$\Delta {H}^{o}$ and

$\Delta {S}^{o}$ for the reaction do not change with temperature, which of the statements is true?
(

$\Delta {H}^{o}=-95kJ$ and

$\Delta {S}^{o}=-190J{ K }^{ -1 }$ )

0%
Ammonia dissociates spontaneously below $500K$
0%
Ammonia dissociates spontaneously above $500K$
0%
Ammonia dissociates at all temperatures
0%
Ammonia does not dissociate at any temperature

Explanation

$\Delta { G }^{ o }=\Delta { H }^{ o }-T\Delta { S }^{ o }$

$=-95\times 1000-500\times (-90)$

$=0$

Thus at temperatures below 500K,

$\Delta G <0$ .

$1g$ sample of substance

$A$ at

${100}^{o}C$ is added to

$100mL$ of

${H}_{2}O$ at

${25}^{o}C$ . Using separate

$100mL$ portion of

${H}_{2}O$ the procedure is repeated with substance

$B$ then with substance

$C$ . How will the final temperatures of the water compare?

Substance	Specific heat
$A$	$0.6{g}^{-1}$ $^{ o }{ { C }^{ -1 } }$
$B$	$0.4{g}^{-1}$ $^{ o }{ { C }^{ -1 } }$
$C$	$0.2{g}^{-1}$ $^{ o }{ { C }^{ -1 } }$

0%
${T}_{C}>{T}_{B}> {T}_{A}$
0%
${T}_{B}>{T}_{A}> {T}_{C}$
0%
${T}_{A}>{T}_{B}> {T}_{C}$
0%
${T}_{A}= {T}_{B}= {T}_{C}$

Explanation

$\text{More the value of Specific heat more the value of final temperature.}$

$T_A>T_B>T_C$

$\Delta H$ and

$\Delta S$ for the reaction

${ Br }_{ 2 }(l)+{ Cl }_{ 2 }(l)\longrightarrow 2BrCl(g)$
are

$29.37\ kJ$ and

$104.0\ J{ K }^{ -1 }$ respectively. Above that temperature will this reaction become spontaneous?

0%
T > 177.8 K
0%
T > 35.1 K
0%
T > 28.4 K
0%
T > 141.2 K

Explanation

According to Gibbs-Helmholtz equation

$\Delta G=\Delta H-T\Delta S$
For spontaneous process,

$\Delta G< 0$
i.e.,

$T\Delta S>\Delta H$

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

$T>\cfrac { 29.37\times 1000 }{ 104 }$

$T> 282.4K$

Following enthalpy changes are given:

$\alpha -D\quad glucose(s)\longrightarrow \alpha -D\quad glucose(aq.);\quad \Delta H=10.72kJ$

$\beta -D\quad glucose(s)\longrightarrow \beta -D\quad glucose(aq.);\quad \Delta H=4.68kJ$

$\alpha -D\quad glucose(aq.)\longrightarrow \beta -D\quad glucose(aq.);\quad \Delta H=1.16kJ$

Calculate the enthalpy change in,

$\alpha -D\quad glucose(s)\longrightarrow \beta -D\quad glucose(s)$

0%
$14.24kJ$
0%
$16.56kJ$
0%
$7.2kJ$
0%
$4.88kJ$

For a particular reaction,

$\Delta { H }^{ o }=-38.3\ kJ$ and

$\Delta { S }^{ o }=-113\ J{ K }^{ -1 }{ mol }^{ -1 }$ . This reaction is:

0%
spontaneous at all temperatures
0%
non-spontaneous at all temperatures
0%
spontaneous at temperatures below ${66}^{o}C$
0%
spontaneous at temperatures above ${66}^{o}C$

Explanation

${ \left( \Delta { G } \right) }=$ ${ \left( \Delta { H } \right) }-T$ ${ \left( \Delta { S } \right) }$

for spontanious ${ \left( \Delta { G } \right) }$ must be negative.

${ \left( \Delta { H } \right) }-T$ ${ \left( \Delta { S } \right) }$ <0

${ \left( \Delta { H } \right) }<T$ ${ \left( \Delta { S } \right) }$

${ \left( \Delta { H } \right) }$ / ${ \left( \Delta { S } \right) }$ < $T$

$(-38300)$ / $(-113)$ < $T$

$339K$ < $T$

$66^oC$ < $T$

Under which circumstances would the free energy change for a reaction be relatively temperature independent?

0%
$\Delta { H }^{ o }$ is negative
0%
$\Delta { H }^{ o }$ is positive
0%
$\Delta { S }^{ o }$ has a large positive value
0%
$\Delta { S }^{ o }$ has a small magnitude

Explanation

${ \left( \Delta { G } \right) }=$

${ \left( \Delta { H } \right) }-$

$T$

${ \left( \Delta { S } \right) }$

$\text{From the above equation we can say for being independence of temperature }$

${ \left( \Delta { S } \right) }$

$\text{ has to become small in magnitude or almost zero.}$

Given that:

$2C(s)+2{ O }_{ 2 }(g)\longrightarrow 2C{ O }_{ 2 }(g);\Delta H=-787kJ...(i)$

${ H }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow { H }_{ 2 }O(l);\Delta H=-286kJ...(ii)$

${ C }_{ 2 }{ H }_{ 2 }(g)+\cfrac { 5 }{ 2 } { O }_{ 2 }(g)\longrightarrow 2C{ O }_{ 2 }(g)+3{ H }_{ 2 }O(l)....(iii)$

$\Delta H=-1301kJ$
Heat formation of acetylene is:

0%
$-1802kJ$
0%
$+1802kJ$
0%
$-800kJ$
0%
$+228kJ$

Explanation

Required equation is:

$2C(s)+{ H }_{ 2 }(g)\longrightarrow { C }_{ 2 }{ H }_{ 2 }(g)$

$eq.(i)+eq.(ii)-eq.(iii)$ gives

$\Delta H=+228kJ$

Which is correct about

$\Delta G$ ?

0%
$\Delta G=\Delta H-T\Delta S$
0%
At equilibrium $\Delta { G }^{ o }=0$
0%
At eq. $\Delta G=-RT\log{K}$
0%
$\Delta G=\Delta { G }^{ o }+RT\log{K}$

Explanation

Gibb's free energy equation :

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

For a reaction,

$\Delta G={ \Delta G }^{ 0 }+RTlnK$

At equilibrium,

$\Delta G=0$

$\therefore \quad { \Delta G }^{ 0 }=-RTlnK$ (at equilibrium)

The enthalpy changes for two reactions are given by the equations:

$2Cr(s)+1\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow { Cr }_{ 2 }{ O }_{ 3 }(s);\Delta H=-1130kJ$

$C(s)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow CO(g);\Delta H=-110kJ$
What is the enthalpy change, in

$kJ$ , for the reaction?

$3C(s)+{ Cr }_{ 2 }{ O }_{ 3 }(s)\longrightarrow 2Cr(s)+3CO(g)$

0%
$-1460kJ$
0%
$-800kJ$
0%
$+800kJ$
0%
$+1020kJ$
0%
$+1460kJ$

Explanation

${ Cr }_{ 2 }{ O }_{ 3 }(s)\longrightarrow 2Cr(s)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g);\Delta H=+1130kJ$

$3C(s)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow 3CO(g);\Delta H=-330kJ$
Adding both equaitons

$\Delta H=+800kJ$

The standard Gibbs free energy

$\Delta { G }^{ o }$ is related to equilibrium constant

${K}_{P}$ as:

0%
${K}_{P}=-RT\log{\Delta { G }^{ o }}$
0%
${K}_{P}={[e/RT]}^{\Delta { G }^{ o }}$
0%
${K}_{P}=-\Delta { G }^{ o }/RT$
0%
${K}_{P}={e}^{-\Delta { G }^{ o }/RT}$

Explanation

For a reaction, the Gibb's free energy equation is :

$\Delta G=\Delta { G }^{ 0 }+RTln{ K }_{ P }$

At equilibrium,

$\Delta G=0$

$\therefore \quad { \Delta G }^{ 0 }=-RTln{ K }_{ P }$

$\therefore \quad { K }_{ P }={ e }^{ -{ \Delta G }^{ 0 }/RT }$

Hess's law of constant heat summation in based on:

0%
$E=mc^{2}$
0%
Conservation of mass
0%
First law of thermodynamics
0%
None of the above

Hess law of heat summation includes

0%
Initial reactants only
0%
Initial reactants and final products
0%
Final products only
0%
Intermediates only

The enthalpy change for two reactions are given by the equations:

$2Cr(s)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow { Cr }_{ 2 }{ O }_{ 3 }(s);\Delta H=-1130kJ$

$C(s)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow CO(g);\Delta H=-110kJ$
What is the enthalpy change in

$kJ$ for the following reaction?

$3C(s)+{ Cr }_{ 2 }{ O }_{ 3 }(s)\longrightarrow 2Cr(s)+3CO(g)$

0%
$-1460kJ$
0%
$-1800kJ$
0%
$+800kJ$
0%
$-1020kJ$
0%
$+1460kJ$

Explanation

$3C(s)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow 3CO(g);\ \Delta H=-330kJ\quad$

${ Cr }_{ 2 }{ O }_{ 3 }(s)\longrightarrow 2Cr(s)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g);\ \Delta H=+1130kJ\quad$

On adding above two reaction:

$3C(s)+{ Cr }_{ 2 }{ O }_{ 3 }(s)\longrightarrow 2Cr(s)+3CO(g);\ \Delta H=-330+1130=800\ kJ$

Given that:

$2Fe(s)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow { Fe }_{ 2 }{ O }_{ 3 }(s);\Delta H=-193.4kJ...(i)$

$Mg(s)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow MgO(s);\Delta H=-140.2kJ....(ii)$
What is

$\Delta H$ of the reaction?

$3Mg+{ Fe }_{ 2 }{ O }_{ 3 }\longrightarrow 3MgO+2Fe$

0%
$-227.2kJ$
0%
$-237.3kJ$
0%
$2227.2kJ$
0%
$-257.3kJ$

Explanation

Eq.

$(ii)$ multiplied by

$3$ ,

$3Mg+{ Fe }_{ 2 }{ O }_{ 3 }\longrightarrow 3MgO;\Delta H=-420.6kJ...(iii)$

$2Fe(s)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow { Fe }_{ 2 }{ O }_{ 3 }(s);\Delta H=-193.4kJ...(i)$
Subtracting

$(i)$ from

$(iii)$

$3Mg+{ Fe }_{ 2 }{ O }_{ 3 }\longrightarrow 3MgO+2Fe\quad \Delta H=-420.6-(-193.4)$

$=-227.2kJ$

A 100.0g ice cube at 0.0°C is placed in 650g of water at 25°C. What is the final temperature of the mixture?

0%
$10^0C$
0%
$11^0C$
0%
$12^0C$
0%
$13^0C$

For spontaneous process:

0%
$\Delta { S }_{ total }=0$
0%
$\Delta { S }_{ total }>0$
0%
$\Delta { S }_{ total }<0$
0%
none of these

Explanation

$\text{Change in entropy is always positive.}$

Reaction of silica with mineral acids may be given as:

$Si{ O }_{ 2 }+4HF\longrightarrow Si{ F }_{ 4 }+2{ H }_{ 2 }O;\Delta H=-10.17kcal$

$Si{ O }_{ 2 }+4HCl\longrightarrow Si{ Cl }_{ 4 }+2{ H }_{ 2 }O;\Delta H=+36.7kcalHCl$
Which among the following is correct?

0%
$HF$ and $HCl$ both will react with silica
0%
Only $HF$ will react with silica
0%
Only $HCl$ will react with silica
0%
Neither $HF$ nor $HCl$ will react with silica

Explanation

When silica reacts with

$HCl$ then the reaction is endothermic and heat is absorbed by the system, when silica reacts with

$HF$ we get

$\Delta H = -ve$ , i.e., exothermic reaction and hence the reaction become feasible,

so, we can say that only

$HF$ reacts with silica.

Calculate the heat of formation of methane, given that
heat of formation of water

$=-286kJ\quad { mol }^{ -1 }$
heat of combustion of methane

$=-890kJ\quad { mol }^{ -1 }$
heat of combustion of carbon

$=-393.5kJ\quad { mol }^{ -1 }$

0%
$75.5\ kJ\ { mol }^{ -1 }$
0%
$-75.5\ kJ\ { mol }^{ -1 }$
0%
$-55.5\ kJ\ { mol }^{ -1 }$
0%
$55.5\ kJ\ { mol }^{ -1 }$

Explanation

Heat of formation of methane = 2

$\times$ heat of formation of water + heat of combustion of carbon - heat of combustion of methane

Heat of formation of methane =

$2*(-286) + (-393.5) -(-890)$ =

$-75.5 KJmol^{-1}$

From the following data of heats of combustion, find the heat of formation of

${ CH }_{ 3 }OH(l)$ :

${ CH }_{ 3 }OH(l)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow C{ O }_{ 2 }(g)+2{ H }_{ 2 }O(l);\Delta H=-726kJ$

$C(s)+{ O }_{ 2 }(g)\longrightarrow C{ O }_{ 2 }(g);\Delta H=-394kJ$

${ H }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow { H }_{ 2 }O(l);\Delta H=-286kJ$

0%
$240\ kJ\ { mol }^{ -1 }$
0%
$-240\ kJ\ { mol }^{ -1 }$
0%
$-140\ kJ\ { mol }^{ -1 }$
0%
$140\ kJ\ { mol }^{ -1 }$

Explanation

Heat of formation of

$CH_3OH$ = 2

$\times$ heat of combustion of

$H_2O$ + heat of combustion of

$CO_2$ - heat of combustion of

$CH_3OH$

Heat of formation of

$CH_3OH$ =

$2*(-286) - (-394) - (-726) = -240 KJmol^{-1}$

The free energy for a reaction having

$\Delta H=31400\ cal,\Delta S=32\ cal\ { K }^{ -1 }{mol}^{-1}$ at

${1000}^{o}C$ is:

0%
$-9336\ cal$
0%
$-7386\ cal$
0%
$-1936\ cal$
0%
$+9336\ cal$

Explanation

We have,

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

$\Delta G = 31400 - 1273*32 = -9336 cal$

The most abundant element in the universe is thought to be:

0%
Hydrogen
0%
Carbon
0%
Oxygen
0%
Nitrogen

A spontaneous process may be defined as :

0%
A process which is exothermic and evolves a lot of heat
0%
A process which is slow and reversible
0%
A process which takes place only in presence of a catalyst
0%
A process that occurs without any input from the surroundings

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 5 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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