MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 6

For the reaction,

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

$\Delta H=29.37kJ\quad { mol }^{ -1 };\Delta S=104J\quad { mol }^{ -1 }$ . Find the temperature above which the reaction would become spontaneous.

0%
Above $273.4\ K$
0%
Above $283.4\ K$
0%
Above $282.4\ K$
0%
Above $373.4\ K$

Explanation

We have,

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

so, spontaneous reaction we have

$\Delta G <0$

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

$29370 - T*104 <0$

so, temperature should be above 282.4 K

Ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance

$100\Omega$ by passing a current of

$1A$ for five minutes, then change in internal energy of the gas is

0%
$0kJ$
0%
$30kJ$
0%
$10kJ$
0%
$20kJ$

Explanation

$\Delta W=0$

$\therefore \Delta Q=\Delta U$

$\Delta Q=\Delta U={ I }^{ 2 }R\Delta t={ 1 }^{ 2 }(100)(5\times 60)=30kJ$

Which of the following is a feasible reaction?

0%
$Ba(s)+{ K }_{ 2 }{ SO }_{ 4 }(aq)\rightarrow Ba{ SO }_{ 4 }(aq+2K(s)$
0%
$Zn(s)+2Ag{ NO }_{ 3 }(aq)\rightarrow Zn{ \left( { NO }_{ 3 } \right) }_{ 2 }(aq)+2Ag(s)\quad$
0%
$Mg(s)+{ Na }_{ 2 }{ SO }_{ 4 }(aq)\rightarrow Mg{ SO }_{ 4 }(aq)+2Na(s)\quad$
0%
$Cu(s)+Mg{ SO }_{ 4 }(aq)\rightarrow Cu{ SO }_{ 4 }(aq)+Mg(s)$

Explanation

$\bf{Hint-}$ A metal can displace the other metal from its salt solution only when the other metal is placed above the first metal in the electrochemical series.

$\bf{Explanation-}$

$\bullet$ Consider the reaction between barium metal and potassium sulphate to form barium sulphate and potassium metal.

$Ba(s) + K_2SO_4(aq) \rightarrow BaSO_4(aq) + 2K(s)$

Barium metal can not displace potassium metal from potassium sulphate, as potassium metal is placed below the barium metal in the electrochemical series.

$\bullet$ Consider the reaction between zinc metal and silver nitrate to form zinc nitrate and silver metal.

$Zn(s) + 2AgNO_3(aq)\rightarrow Zn(NO_3)_2(aq) + 2Ag(s)$

Zinc metal can displace silver metal from silver nitrate, as silver metal is placed above the zinc metal in the electrochemical series.

$\bullet$ Consider the reaction between magnesium metal and sodium sulphate to form magnesium sulphate and sodium metal.

$Mg(s) + Na_2SO_4(aq)\rightarrow MgSO_4(aq) + 2Na(s)$

Magnesium metal can not displace sodium metal from sodium sulphate, as sodium metal is placed below the magnesium metal in the electrochemical series.

$\bullet$ Consider the reaction between copper metal and magnesium sulphate to form copper sulphate and magnesium metal.

$Cu(s) + MgSO_4(aq)\rightarrow CuSO_4(aq) + Mg(s)$

Copper metal can not displace magnesium metal from magnesium sulphate, as magnesium metal is placed below the copper metal in the electrochemical series.
Electrochemical Series: Examples & Uses - Video & Lesson Transcript | Study.com

Electrochemical Series: Examples & Uses - Video & Lesson Transcript | Study.com

$\bf{Conclusion-}$ Hence, option B is correct which is a feasible reaction.

Heat of combustion

$\Delta H^o$ for

$C(s), H_2(g)$ and

$CH_4(g)$ are

$-94,-68$ and

$-213 kcal/mol$ respectively. Then

$\Delta H^o$ for

$C(s) + 2H_2(g) \rightarrow CH_4$ is:

0%
$-17\, kcal/mol$
0%
$-111\, kcal/mol$
0%
$-170\, kcal/mol$
0%
$-85\, kcal/mol$

Explanation

For reaction,

$C(s)+2H_2(g)\rightarrow CH_4(g), \Delta H^o=?$

$\Delta H^o$ =

$-$ [(

$\Delta H^o$ of combustion of

$CH_4$ ) - (

$\Delta H^o$ of combustion of

$C$ )+ (2

$\times \Delta H^o$ of combustion of

$H_2$ )]

$C + O_2 \rightarrow CO_2; \Delta H^o = - 94\, kcal$ ... (i)

$2H_2+O_2 \rightarrow 2H_2O; \Delta H^o = - 68\times 2\, kcal$ ... (ii)

$=-[(-213)-(-94+2\times -68)]\, kcal/mol$

$=-[-213+230]=-17\,kcal/mol$

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?

0%
Exothermic and decreasing disorder
0%
Endothermic and increasing disorder
0%
Exothermic and increasing disorder
0%
Endothermic and decreasing disorder

Explanation

A chemical reaction which is exothermic and has an increasing disorder is certain to result in a spontaneous reaction.
Since the reaction is exothermic,

$\displaystyle \Delta H$ is negative. Since, the disorder increases,

$\displaystyle \Delta S$ is positve.

$\displaystyle \Delta G =\Delta H - T\Delta S = -ve - ( +ve \times +ve) =-ve$

Negative value of

$\displaystyle \Delta G$ corresponds to spontaneous process.

Heat is supplied to a certain homogeneous sample of matter, at a uniform rate. Its temperature is plotted against time, as shown. Which of the following conclusions can be drawn? .

0%
Its specific heat capacity is greater in the solid state than in the liquid state.
0%
Its specific heat capacity is smaller in the solid state than in the liquid state.
0%
Its latent heat of vaporization is greater than its latent heat of fusion.
0%
Its latent heat of vaporization is smaller than its latent heat of fusion.

Explanation

The horizontal part of the curve ,where the system absorbs heat at constant temperature must depict changes of state.Here the latent heats are proportional to length of the horizontal part. In the loping parts specific heat capacity is inversely proportional to the slopes.

Option

$C$ Its latent heat of vaporization is greater then its latent heat of fusion.

Match the List-II ans List-II and List III:

List-I	List-II	List-III
A. $\Delta \, G>0$	X. $\Delta \, S>0$	Non-spontaneous
B. $\Delta \, G<0$	Y. $\Delta \, S<0$	Spontaneous
C. $\Delta \, G=0$	Z. $\Delta \, S=0$	Equilibrium

select the correct answering from the following codes:

0%
A-(Y, 1) B-(X, Y) C-(Z, 3)
0%
A-(X, 2) B-(Y, 3) C-(Z, 1)
0%
A-(X, 3) B-(Y, 1) C-(Z, 2)
0%
A-(Y, 1) B-(X, 3) C-(Z, 2)

Explanation

According to the third law, the entropy change of a spontaneous process is always greater than zero.

For spontaneous process

$\Longrightarrow \quad \Delta G<0\quad and\quad \Delta S>0$

For non-spontaneous process

$\Longrightarrow \quad \Delta G>0\quad and\quad \Delta S<0$

For equilibrium process

$\Longrightarrow \quad \Delta G=0\quad and\quad \Delta S=0$

One mole of a real gas is subjected to a process from

$(2\ bar, 30\ lit, 300\ K)$ to

$(2\ bar, 50\ lit, 400\ K)$
Given:

$C_{V} = 40\ J/mol\ K$ ,

$C_{P} = 50\ J/mol/ K$
Calculate

$\Delta U$ .

0%
$4000\ J$
0%
$2000\ J$
0%
$1000\ J$
0%
$5000\ J$

Explanation

Change in internal energy is written as:

$\Delta U = nC_v \Delta T$

$\Delta U = 1\times 40\times (400-300)$

$\Delta U = 4000J$

Consider the reaction at

$300\ K$ ,

$H_{2}(g) + Cl_{2}(g)\rightarrow 2HCl(g), \ \ \ \Delta H^{\circ} = -185\ kJ$

$3$ mole of

$H_{2}$ completely reacts with

$3$ mole of

$Cl_{2}$ to form

$HCl$ , What is

$\Delta U^{\circ}$ for this reaction?

0%
$0$
0%
$-185\ kJ$
0%
$555\ kJ$
0%
None of these

Explanation

$\begin{array}{l} H_{ 3 }^{ + }\, \, \, C{ l_{ 2 } }\, \, \, \to 2HCL,\, \, \, \, \Delta { H^{ \circ } }=-18kJ \\ 1mol\, \, \, \, \, 1\, mol\, \, \, \, \, \, \, \, \, \, \, \, 2mol \\ 3mol\, \, \, \, \, 3mol\, \, \, \, \, \, \, \, \, \, \, \, 6mol \\ \Delta ng=0 \\ \Delta { H^{ \circ } }=\Delta { U^{ 6 } } \\ \Rightarrow \Delta { U^{ 6 } }=\Delta { H^{ \circ } }=-185KJ. \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$

${ H }_{ 2 }\left( g \right) +\dfrac { 1 }{ 2 } { O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right)$

${ H }_{ 2 }O\left( l \right) \longrightarrow { H }_{ 2 }O\left( g \right); \Delta H={ x }_{ 4 }$
Given,

${ E }_{ H-H }={ x }_{ 1 }$

${ E }_{ O-O }={ x }_{ 2 }$

${ E }_{ O-H }={ x }_{ 3 }$

$\Delta { H }_{ F }$ of

${ H }_{ 2 }O$ vapour is:

0%
${ x }_{ 1 }+\dfrac { { x }_{ 2 } }{ 2 } -{ x }_{ 3 }+{ x }_{ 4 }$
0%
$2{ x }_{ 3 }-{ x }_{ 1 }-\dfrac { { x }_{ 2 } }{ 2 } -{ x }_{ 4 }$
0%
${ x }_{ 1 }+\dfrac { { x }_{ 2 } }{ 2 } -2{ x }_{ 3 }-{ x }_{ 4 }$
0%
${ x }_{ 1 }+\dfrac { { x }_{ 2 } }{ 2 } -2{ x }_{ 3 }+{ x }_{ 4 }$

Explanation

${ H }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right) \quad ;\quad \Delta H$

$\Rightarrow \Delta H=2\left( 0-H \right) -\left( H-H \right) -\dfrac { 1 }{ 2 } \left( 0-0 \right)$

$=2{ x }_{ 3 }-{ x }_{ 1 }-\dfrac { { x }_{ 2 } }{ 2 }$

for

${ H }_{ 2 }O\left( l \right) \longrightarrow { H }_{ 2 }O\left( g \right) \quad \Delta H={ x }_{ 4 }$

${ \Delta H }_{ f }={ x }_{ 4 }-\left[ { 2x }_{ 3 }-{ x }_{ 1 }-\dfrac { { x }_{ 2 } }{ 2 } \right]$

$={ x }_{ 1 }+\dfrac { { x }_{ 2 } }{ 2 } -2{ x }_{ 3 }+{ x }_{ 4 }$

Ans :- D

What is the heat produced when 4 g of anhydrous solid is added to 50 g of water?

0%
400 kJ
0%
1672 J
0%
200 kJ
0%
836 kJ

Explanation

As it is mention in the that either the temperature is rising or not we will take it as constant then the heat produced will be

$Heat\ produced=50\times4=200kJ$

Hence 200kJ heat is produced when 4g of anhydrous solid is added to 50g of water.

$\triangle H_{vaporisation}$ of substance

$X(l)$ (molar mass:

$30\ g/mol)$ is

$300\ J/g$ at it's boiling point

$300\ K$ , then molar entropy change for reversible condensation process is____________.

0%
$30\ J/mol.K$
0%
$-300\ J/mol.K$
0%
$-30\ J/mol.K$
0%
None of these

Explanation

Solution:- (A)

$30 \; {J}/{mol-K}$

Given:-

${\Delta{H}}_{vap.} = 300 \; {J}/{g}$

Molar mass of

$X = 30 \; {g}/{mol}$

$\therefore {\Delta{H}}_{molar} = 300 \times 30 = 9000 \; {J}/{mol}$

Boiling point of substance

$X, \; \left( T \right) = 300 K$

As we know that,

Molar entropy change,

${\Delta{S}}_{molar} = \cfrac{{\Delta{H}}_{molar}}{T} = \cfrac{9000}{300} = 30 \; {J}/{mol-K}$

Hence the molar entropy change for the given process will be

$30 \; {J}/{mol-K}$ .

$27 C$ the reaction,

${ C }_{ 6 }{ H }_{ 6\left( g \right) }\ +\ { 15 }/{ 2\ { O }_{ 2\left( g \right) } }\longrightarrow 6C{ O }_{ 2\left( g \right) }\ +\ 3{ H }_{ 2 }{ O }_{ \left( l \right) }$
proceeds spontaneously because the magnitude of__________.

0%
$\Delta H=T\Delta S$
0%
$\Delta H>T\Delta S$
0%
$\Delta H < T \Delta S$
0%
$\Delta H>0$ and $T\Delta S<0$

The equilibrium constant of a reaction at

$298K$ is

$5\times {10}^{-3}$ and at

$1000K$ is

$2\times {10}^{-5}$ . What is the sign of

$\Delta H$ for the reaction?

0%
$\Delta H$ is +ve
0%
$\Delta H$ is -ve
0%
$\Delta H\propto 0$
0%
$\Delta H$ is $\pm$ ve

Explanation

Relation between equilibrium constant and

$\Delta H$ is as follows-

$\log \dfrac{K_2}{K_1} = \dfrac{\Delta_H}{2.303R} [\dfrac{ 1}{T_1} - \dfrac{1}{T_2}]$

$\log \dfrac {2\times 10^{-5}}{5\times 10^{-3}} = \dfrac{\Delta H}{2.303R} \dfrac{1}{T_1} -\dfrac{1}{T_2}$

$\log 4\times 10^{-3} = \dfrac{\Delta H}{2.303R}\times 2.3\times 10^{-3}$

So Sign of

$\Delta H$ would be positive.

$\Delta$ G is the net energy available to do useful work and is a measure of free energy. If a reaction has positive enthalpy change and positive entropy change, under what conditions will the reaction be spontaneous?

0%
$\Delta$ G will be positive at low temperature hence reaction is spontaneous at low temperature.
0%
$\Delta$ G will be negative at high temperature hence reaction is spontaneous at high temperature.
0%
$\Delta$ G will be negative at low temperature hence reaction is spontaneous at low temperature.
0%
$\Delta$ G will be negative at all temperature hence reaction is spontaneous at all temperatures.

Explanation

$\Delta H>0$ and

$\Delta S>0$

Now,

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

For a spontaneous process,

$\Delta G<0,i.e, T\Delta S > \Delta H$ , i.e., temperature should be high.

Therefore, the reaction will be spontaneous for negative value of

$\Delta G$ , i.e., high temperature is required.

$H_{2(g)} + \dfrac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}$

$\Delta H = 241.8 kJ$

$CO_{(g)} + \dfrac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$

$\Delta H = 283 kJ$

The heat evolved during the combustion of

$112$ litre of water gas (mixture of equal volume of

${H}_{2}$ and

$CO$ ) is:

0%
$241.8kJ$
0%
$283kJ$
0%
$1312kJ$
0%
$1586kJ$

Explanation

Solution:- (C)

$1312 \; KJ$

Molecular formula of water gas

$= {H}_{2} + CO$

Combustion of

${H}_{2}$ -

${H}_{2} + \cfrac{1}{2} {O}_{2} \longrightarrow {H}_{2}O; \; \Delta{H} = 241.3 \; KJ$

Combustion of

$CO$ -

$CO + \cfrac{1}{2} {O}_{2} \longrightarrow C{O}_{2}; \; \Delta{H} = 283 \; KJ$

$1$ mole of

${H}_{2}$ and

$CO$ each gives,

$\Delta{H} = 283 + \left( 241.8 \right) = 524.8 \; KJ$

$1$ mole of

${H}_{2} = 22.4 \; L$

Similarly,

$1$ mole of

$CO = 22.4 \; L$

$\therefore$ Total volume

$= 22.4 + 22.4 = 44.8 \; L$

Therefore,

Heat evolved during combustion of

$44.8 \; L$ of water gas

$= 524.8 \; KJ$

Heat evolved during combustion of

$112 \; L$ of water gas

$= \cfrac{524.8 \times 112}{44.8} = 1312 \; KJ$

Hence the heat evolved during the combustion of

$112$ litre of water gas at STP is

$1312 \; KJ$ .

The specific heat of a gas is found to be

$0.075$ calories at constant volume and its formula weight is

$40$ . The atomicity of the gas would be:

0%
one
0%
two
0%
three
0%
four

Explanation

Total Heat

$=40\times 0.075=3$ cal

$=Cv$

$Cp-Cv=R$

$Cp=Cv+R$

$=3+2$

$=5$ cal

$\dfrac{Cp}{Cv}=\dfrac{5}{3}$

$=$ monoatomic.

1165876_827145_ans_69b1671541eb4075aec1a6bb921e3101.jpg

A piston cylinder device initially contains

$0.2m^3$ neon (assume ideal) at 200 kPa inside at

$T_1 ^0C$ . A value is now opened and neon is allowed to escape until the volume reduces to half the initial volume. At the same time heat transfer with outside at

$T_2 ^0C$ ensures a constant temperature inside. Select the statement(s) for given process :

0%
$\Delta U$ must be zero
0%
$\Delta U$ may not be zero
0%
q may be +ve
0%
q may not be -ve

Explanation

The process is Isothermal as temperature constant inside.

for Isothermal

$DU=0$ when gas expands work done by the system is

$-ve$ and q is positive.

Option A is correct.

Hess's law is applicable for the determination of heat of ________.

0%
transition
0%
formation
0%
reaction
0%
all of these.

Explanation

Hess's Law of Constant Heat Summation: The standard enthalpy of a reaction, which takes place in several steps, is the sum of the standard enthalpies of intermediate reactions into which the overall reactions may be divided at the same temperature.

$\Delta H=\Delta { H }_{ 1 }+\Delta { H }_{ 2 }+\Delta { H }_{ 3 }$

It is applicable to determine heat of formation, transition, reaction, hydration, etc.

$Al^{3+} (aq) + 3e^{-}\rightarrow Al(s); E^{\circ} = -1.66\ V$

$Cu^{2+} (aq) + 2e^{-}\rightarrow Cu(s); E^{\circ} = + 0.34\ V$
What voltage is produced under standard conditions to give a spontaneous reactions by combination of these two-half cells?

0%
$1.32\ V$
0%
$-1.32\ v$
0%
$2.00\ V$
0%
$-2.00\ V$

Explanation

${ Al }^{ 3+ }+3{ e }^{ - }\rightarrow Al\quad { E }^{ 0 }=-1.66$

${ Cu }^{ 2+ }+{ 2e }^{ - }\rightarrow Cu\left( s \right) \quad ;\quad { E }^{ 0 }=+0.34V$

$\Rightarrow$ When

${ 1 }^{ st }$ reaction is reversed

$\Rightarrow$

$Al\longrightarrow { Al }^{ 3+ }+3{ e }^{ - }\Rightarrow { E }^{ 0 }=1.66V$

$\Rightarrow$

$Al$ gets reduced and

${ Cu }^{ 2+ }$ gets oxidized.

So,

${ E }_{ 1 }-{ E }_{ 2 }=1.66V-0.34V=1.32V$

Supposing the distance between the atoms of a diatomic gas to be constant, its specific heat at constant volume per mole(gram mole) is?

0%
$\frac{5}{2}R$
0%
$\frac{3}{2}R$
0%
R
0%
$\frac{7}{2}R$

Explanation

For diatomic gas specific heat of constant volume

$C_{V}=\dfrac{5}{2}R$

Which of the following statement(s) is/are true?
"Internal energy of an ideal gas ............"

0%
Decreases in an isobaric process
0%
Remains constant in an isothermal process
0%
Increases in an isobaric process
0%
Decreases in an isobaric expansion

Explanation

In isothermal process

$\Delta U=0$
In isobaric expansion

$V\propto T$ so

$\Delta$ U increases.

A reaction proceeds through two paths I and II to convert

$X$

$\rightarrow$

$Z$ . What is the correct relationship between

$Q$ ,

$Q_1$ and

$Q_2$ (

$Q$ represents a change in internal energy, here) ?

0%
$Q = Q_1 \times Q_2$
0%
$Q = Q_1 + Q_2$
0%
$Q = Q_2 - Q_1$
0%
$Q = Q_1 / Q_2$

Explanation

Internal energy (Q) is a state function.

$\therefore { Q }_{ XZ }={ Q }_{ XY }+{ Q }_{ YZ }$

$\therefore Q={ Q }_{ 1 }+{ Q }_{ 2 }$

Hence, the correct answer is option

$\text{B}$ .

The value of

$\Delta E$ for combustion of 16 g of

$CH_4$ is -885389 J at 298 K. The

$\Delta H$ combustion for

$CH_4$ in J

$mol^{-1}$ at this temperature will be:

$(Given that, R \, = \, 8.314 \, JK^{-1} \, mol^{-1})$

0%
$-55337$
0%
$-880430$
0%
$-885389$
0%
$-890348$

Explanation

$\triangle H=\triangle E+\triangle n_gRT$

$\underset{(g)}{CH_4}+\underset{(g)}{2O_2}\longrightarrow \underset{(g)}{CO_2}+\underset{(g)}{2H_2O}$

Since all are attained in gaseous form and

$\triangle H_{condensation }$ of

$H_2O$ is not provided

$\therefore \triangle n_g$ =no of moles of gaseous reactants

$-$ no of moles of gaseous products

$=3-3=0$

$\therefore \Delta H=\Delta E+ORT$

$\therefore \Delta H_{combustion}=-885389\ Joules$

The statement "The change of enthalpy of a chemical reaction is same whether the reaction takes place in one or several steps" is:

0%
Le Chatelier's law
0%
van't Hoff's law
0%
first law of thermodynamics
0%
Hess's law.

$C_2H_6(g) + \frac{7}{2}O_2(g) \rightarrow 2CO_2 (g) + 3H_2O(l)$

$\delta E - \delta H$ for this reaction

$27^0$ C will be :

0%
$+ 1247.1$ J
0%
$- 1247.1$ J
0%
$-6235.5$ J
0%
$+6235.5$ J

Explanation

Solution:- (D)

$+6235.5 \; J$

${{C}_{2}{H}_{6}}_{\left( g \right)} + \cfrac{7}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow 2 {C{O}_{2}}_{\left( g \right)} + 2 {{H}_{2}O}_{\left( l \right)}$

$\Delta{{n}_{g}}$ for the above reaction-

$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = \left( 2 + 0 \right) - \left( 1 + \cfrac{7}{2} \right) = - \cfrac{5}{2}$

$T = 27 ℃ = \left( 27 + 273 \right) = 300 K \; \left( \text{Given} \right)$

Now, from first law of thermodynamics,

$\Delta{H} = \Delta{E} + \Delta{{n}_{g}} RT$

$\Rightarrow \Delta{E} - \Delta{H} = - \Delta{{n}_{g}} RT = - \left[ \left( -\cfrac{5}{2} \right) \times 8.314 \times 300 \right] = +6235.5 \; J$

Hence the value of

$\Delta{E} - \Delta{H}$ for the given reaction is

$+6235.5 \; J$

For the reaction at

$\ { 25 }^{ o }C,{ X }_{ 2 }{ O }_{ 4(l) }\longrightarrow 2X{ O }_{ 2(g) }\quad$

$\Delta H=2.1kcal$ and

$\Delta S=20cal\quad { K }^{ -1 }$ .

The reaction would be:

0%
spontaneous
0%
non-spontaneous
0%
at equilibrium
0%
unpredictable

Explanation

Answer:- (A) Spontaneous

Given:-

$= 25^oC = (273 + 25)K = 298K$

$\Delta H = 2.1 kcal = 2.1 \times {10}^{3} cal \; \& \; \Delta S = 20cal{K}^{-1}$

Gibbs free energy

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

$\Delta G = 2.1 \times {10}^{3} - (298 \times 20) = 2100 - 5960 = -2860cal$

$\because \Delta G = negative$ , thus the reaction is spontaneous.

The total entropy change (

$\Delta S_{total}$ ) for the system and surroundings of a spontaneous process is given by :

0%
$\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} > 0$
0%
$\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} < 0$
0%
$\Delta S_{system} = \Delta S_{total} + \Delta S_{surr} > 0$
0%
$\Delta S_{surr} = \Delta S_{total} + \Delta S_{system} < 0$

For a reaction to be spontaneous at any temperature, the conditions are:

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

Explanation

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

$\Delta H$ = negative and

$\Delta S$ is positive , then for any value of T ,

$\Delta G$ will be negative and the process will be spontaneous.

Which of the following statements is not correct?

0%
For a spontaneous process, $\Delta G$ must be negative.
0%
Enthalpy, entropy, free energy etc. are state variables.
0%
A spontaneous process is reversible in nature.
0%
Total of all possible kinds of energy of a system is called its internal energy.

Explanation

For a spontaneous process,

$\Delta G<0$

The variables which depend only on the initial and final states of the system are state variables.

Therefore, enthalpy, entropy and free energy are state variables.

A spontaneous process is not reversible in nature as for opposite reaction to occur, a force has to be applied.

The internal energy is the total of all possible kinds of energy in the system.

The volume of gas is reduced to half from its original volume. The specific heat will:

0%
get reduced to half
0%
get doubled
0%
remains constant
0%
get increased by four times

Explanation

Hint: Specific heat is an intensive property.

Correct Answer: Option (C).

Explanation:

Specific heat is defined as the amount of heat required to increase the temperature of one unit mass of a substance by $1^◦C$ .

It is also an intensive property. So, an increase or decrease in volume can’t affect specific heat. It remains constant throughout the system.

Final Answer: The volume of gas is reduced to half from its original volume. The specific heat will remain constant.

1 mole of an ideal gas STP is subjected to a reversible adiabatic expansion to double its volume. The change in internal energy

$(\gamma = 1.4)$

0%
1169 J
0%
769 J
0%
1373 J
0%
969 J

Explanation

Given,

$\gamma=1.4$

$n=1$

At STP,

$T_1=273.15K$

$V_1=V$

$V_2=2V$

In adiabatic process,

$PV^{\gamma}=constant=k$ . . . . . . . .(1)

From ideal gas equation,

$PV=nRT$

$P=\dfrac{1\times RT}{V}=\dfrac{RT}{V}$ . . . . .. (2)

Substitute equation (2) in equation (1), we get

$\dfrac{RT}{V}V^{\gamma}=k$

$TV^{\gamma-1}=k'(constant)$

$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$

$T_2=T_1(\dfrac{V_1}{V_2})^{\gamma-1}$

$T_2=273.15\times (\dfrac{V}{2V})^{1.4-1}$

$T_2=207.00K$

Change in internal energy for adibatic process,

$\Delta U=\dfrac{R}{\gamma-1}(T_1-T_2)$

$\Delta U=\dfrac{8.31}{1.4-1}(273.15-207.00)$

$\Delta U=1373J$

The correct option is C.

Internal energy of an ideal gas depends upon

0%
temperature only
0%
volume only
0%
both volume and temperature
0%
neither volume nor temperature

Explanation

Hint: $U=nC_v\delta T$
Explanation: The internal energy of a gas is an inherent property. For an ideal gas, there is the absence of inter-molecular collision. So, gas has only translational kinetic energy. So, internal energy doesn’t depend on external factors like volume and pressure. So, the internal energy of an ideal gas depends on temperature only.

$Answer:$

Hence, option A is the correct option.

Three stars A, B, C have surface temperatures

$T_A$ ,

$T_B$ and

$T_C$ . A appears bluish, B appears reddish and C appears yellowish. We can conclude that:

0%
$T_A > T_C > T_B$
0%
$T_A > T_B > T_C$
0%
$T_B > T_C > T_A$
0%
$T_C > T_B > T_A$

Explanation

According to Wien's Displacement law we know that:-

$\lambda=\dfrac bT \ldots \ldots (1)$ .

Given that three stars

$A, B$ and

$C$ where

$A$ appears bluish,

$B$ appears reddish and

$C$ appears yellowish.

We know that

$\lambda_{blue}< \lambda_{yellow}< \lambda_{Red}$

$\implies \lambda_{A}<\lambda_{C}<\lambda_{B} ..... (2)$

from (1) and (2) we get

$\implies T_{A} > T_{C}> T_{B}$

Hence option

$(A)$ is correct

The signs of

$\Delta$ H,

$\Delta$ S and

$\Delta$ G for a non-spontaneous reaction at all temperatures would be:

0%
+,+,-
0%
+,-,+
0%
-,-,-
0%
+,+, +

Explanation

Answer:-

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

For a non-spontaneous reaction

$\rightarrow$

$\Delta{G}$ must be positive i.e.

$\Delta{H} \; \& \; \Delta{S}$ must be positive and negative respectively.

$\Rightarrow$

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

$\because \; \Delta{S} = negative$

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

$\Rightarrow \Delta{G} = \Delta{H} + T \Delta{S} = positive$

Hence, B is the correct answer.

1 kg of water is heated from 40 C to 70 C, If its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J

${ kg }^{ -1 }{ K }^{ -1 })$

0%
$2.44\times { 10 }^{ 5 }$ J
0%
$1.62\times { 10 }^{ 5 }$ J
0%
$1.24\times { 10 }^{ 5 }$ J
0%
$2.62\times { 10 }^{ 5 }$ J

Explanation

Since volume does not change as the energy supplied is used to change the internal energy.

$\Delta E=\Delta U=mC\theta\\ \quad=1\times4148\times30\\ \quad=1.24\times 10^5 J$

An electric heater supplies heat to a system at a rate of 120 W. If system performs work at a rate of 80 J

${ s }^{ -1 }$ , the rate of increase in internal energy is :

0%
30 J ${ s }^{ -1 }$
0%
40 J ${ s }^{ -1 }$
0%
50 J ${ s }^{ -1 }$
0%
60 J ${ s }^{ -1 }$

Explanation

Q=120 W

W=80 Js

$^{-1}$

From the first law of thermodynamics:

$Q=U+W$

where U=internal energy

$\therefore U=Q-W =120-80W=40 J s^{-1}$

An ideal gas undergoing a change of state from A to B through four different paths I, II ,III and IV as shown in the P-V diagram that lead to the same change of state, then the change in internal energy is

0%
Same in I and II but not in III and IV
0%
Same in III and IV but not in I and II
0%
Same in I,II and III but not in IV
0%
Same in all of the four cases

The signs of

$\triangle$ H,

$\triangle$ S and

$\triangle$ G for a non-spontaneous reaction at all temperatures would be:

0%
+, +, -
0%
+, -, +
0%
-, -, -
0%
+, +, +

Explanation

$\Delta H<0,\ \Delta S>0\rightarrow$ Spontaneous at any temperature.

$\Delta H>0,\ \Delta S<0\rightarrow$ Non-spontaneous regardless of the temperature.

$\Delta H>0,\ \Delta S>0\rightarrow$ Spontaneous at certain temperature range.

$\Delta H<0,\ \Delta S<0\rightarrow$ Spontaneous at certain temperature range.

$\Delta G<0\ \rightarrow$ In order for reaction to be Spontaneous at given temperature.

$\Delta G>0\ \rightarrow$ In order for reaction to be Non-Spontaneous at given temperature.

Two bars of same length and same cross -sectional area but of different thermal conductivities

$K_1$ and

$K_2$ are joined end to end as shown in the figure .One end of the compound bar is at temperature

$T_1$ and the opposite end at temperature

$T_2$

$(where T_1 > T_2)$ .
The temperature of the junction is

0%
$\dfrac{K_1T_1+K_2T_2}{K_1+K_2}$
0%
$\dfrac{K_1T_2+K_2T_1}{K_1+K_2}$
0%
$\dfrac{K_1(T_1+T_2)}{K_2}$
0%
$\dfrac{K_2(T_1+T_2)}{K_1}$

Explanation

Let

$L$ and

$A$ be length and area of cross-section of each bar respectively.

$\therefore$ Heat current through the bar 1 is

$H_1=\dfrac{K_1A(T_1-T_0)}{L}$

At steady state,

$H_1=H_2$

$\therefore \dfrac{K_1A(T_1-T_0)}{L}=\dfrac{K_2A(T_0-T_2)}{L}$

$K_1(T_1-T_0)=K_2(T_0-T_2)$

$K_1T_1-K_1T_0=K_2T_0-K_2T_2$

$K_1T_0+K_2T_0=K_1T_1+K_2T_2$

$T_0(K_1+K_2)=K_1T_1+K_2T_2$

$T_0=\dfrac{K_1T_1+K_2T_2}{(K_1+K_2)}$

A certain mass of gas is taken from an initial thermodynamic state A to another state B by process I and II. In process I the gas does 5 joules of work and absorbs 4 joules of heat energy. In process II, the gas absorbs 5 joules of heat. The work done by the gas in process II (see figure) is

0%
+6 joules
0%
-6 joules
0%
+4 joules
0%
-4 joules

Explanation

Internal energy doesnt depend on path.

$\therefore Q=U+W$

in path

$I$

$Q=u, w=+5$

$U=-1\ J$

in path

$II$

$Q=5, U=-1$

$\therefore W=5-(-1)=6\ J$

For the combustion of

$CH_4$ at 1 atm pressure & 300 K, which of the following options is correct?

$2M + O_2 \xrightarrow 2MO$

$\Delta G_{1000 ^{\circ}C} = -921 kJ/mol$

$\Delta G_{1900 ^{\circ}C} = -300 kJ/mol$

$2C + O_2 \xrightarrow 2CO$

$\Delta G _{1000 ^{\circ}C} = -432 kJ/mol$

$\Delta G_{1900 ^{\circ}C} = -624 kJ/mol$

$MO + C \xrightarrow{\Delta} M + CO \uparrow$
This reaction is feasible at temperature:

0%
$1900^\circ C$
0%
$1000^\circ C$
0%
$900^\circ C$
0%
$1200^\circ C$

Explanation

$At\quad 100℃,\quad 2M+{ O }_{ 2 }\longrightarrow 2MO;\Delta G=-921kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \therefore 2MO\longrightarrow 2M+{ O }_{ 2 };\Delta G=921kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad MO\longrightarrow M+\cfrac { 1 }{ 2 } { O }_{ 2 };\Delta G=460.5kJ/mole-------(i)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 2C+{ O }_{ 2 }\longrightarrow 2CO;\Delta G=-432kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad C+\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow CO;\Delta G=-216kJ/mole--------(ii)\\ Adding\quad (i)\& (ii),\quad MO+C\xrightarrow { \quad \quad \quad \quad \Delta \quad \quad \quad } M+CO\uparrow ;\Delta G=460.5-216=244.5kJ/mole\\ \therefore \Delta G>0$

$\therefore$ It is not feasible.

$At\quad 1900℃,\quad 2M+{ O }_{ 2 }\longrightarrow 2MO;\Delta G=-300kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \therefore 2MO\longrightarrow 2M+{ O }_{ 2 };\Delta G=300kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad MO\longrightarrow M+\cfrac { 1 }{ 2 } { O }_{ 2 };\Delta G=150kJ/mole-------(iii)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 2C+{ O }_{ 2 }\longrightarrow 2CO;\Delta G=-624kJ/mole\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad C+\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow CO;\Delta G=-312kJ/mole--------(iv)\\ Adding\quad (iii)\& (iv),\quad MO+C\xrightarrow { \quad \quad \quad \quad \Delta \quad \quad \quad } M+CO\uparrow ;\Delta G=150-312=-162kJ/mole\\ \therefore \Delta G<0$

$\therefore$ It is feasible.

Which of the following parameters does not charaterize the thermodynamic state of matter?

0%
temperature
0%
pressure
0%
work
0%
volume

The internal energy

$U$ is a unique function of any state because change in

$U$ .

0%
Does not depends upon path
0%
Depends upon path
0%
Corresponds to adiabatic process
0%
Corresponds to an isothermal process

Explanation

Internal energy of a body depends upon the final and initial temperature of the body. It does not depend upon the path an the process the body has undergoes

So, option

$(A)$

One end of a

$0.25\ m$ long metal bar is in steam and the other end is in contact with ice. If

$12\ g$ of ice melts per minute, what is the thermal conductivity of the metal? Given cross-section of the bar

$= 5\times 10^{-4} m^{2}$ and latent heat of ice is

$80\ cal/g$ .

0%
$80\ cal/s-m-^{\circ}C$
0%
$90\ cal/s-m-^{\circ}C$
0%
$70\ cal/s-m-^{\circ}C$
0%
$60\ cal/s-m-^{\circ}C$

If for a reaction

${k_c}$ = 1 then

$\Delta G^\circ$ =________.

0%
1.987
0%
4.184
0%
0
0%
1

Explanation

$\Delta G=\Delta { G }^{ \circ }+RT㏑K$

At equilibrium,

$\Delta G=0$

$\therefore \Delta { G }^{ \circ }=-RT㏑{ K }_{ C }=-RT㏑1=-RT(0)=0$

"Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement of the consequence of

0%
second law of thermodynamics
0%
conservation of momentum
0%
conservation of mass
0%
first law of thermodynamics

Explanation

Hint: The second law of thermodynamics dictates that an isolated system’s entropy will not decrease with time. And first law states that energy cannot be created not be destroyed and that it is simply transferred from one form to the other

Step 1: Explanation:

The Clausius statement on Second law of thermodynamics states that, “Heat cannot flow from a cold body to a hot body without the performance of work by some external agency.” This is in direct correlation to the second law.

$\textbf{Hence, the correct option is (A)}$

Select the incorrect option about spontaneous exothermic reaction.

0%
$\Delta S_{Surr} = + ve$
0%
$\Delta S_{Overall} = + ve$
0%
$\Delta S_{System} = - ve$
0%
$\Delta S_{System}= + ve$ always

Explanation

Spontaneous reactions will -ve

$\Delta$ G [Gibbs free energy]

And exothermic reaction means -ve

$\Delta$ H

From Formula

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

To maintain -

$\Delta$ G overall

$\Delta$ S should be +ve so option B is correct.

As by the exothermic process we are decreasing the entropy (energy) of a system by increasing entropy of surroundings.

Hence

$\Delta S_{surr}$ = +ve and

$\Delta S_{system}$ may =-ve are correct.

The last one

$\Delta S_{system}$ = +ve always is incorrect by considering the above three statements.

The internal energy of a system remains constant when it undergoes

0%
Cyclic process
0%
An isothermal process
0%
Any process in which the heat given out by the system is equal to work done on the system
0%
All of the above

Which of the following options is correct regarding spontaneity of a process occurring on a system in which only pressure-volume? work is involved and S, G, Cl, H, V, and P have usual meaning as in thermodynamics?

0%
$(dG)_{U, V} < 0, (dS)_{T, V} > 0$
0%
$(dH)_{S, V} < 0, (dG)_{T, P} < 0$
0%
$(dU)_{S, V} < 0, (dG)_{T, V} < 0$
0%
$(dS)_{H, P} > 0, (dG)_{T, P} < 0$

Explanation

For the reaction to be spontenaous

$\Delta G$ that is gibb's free enrgy should be negative and entropy

$(\Delta S)$ should be positive.

For reaction to be spontaneous reaction

$(\Delta H)$ should also be negative i.e exothermic reaction.

Mathematically

$(dS)_{H.P}>0$

$(dG)_{T,G}<0$

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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