$$\Delta G$$ increases and the reaction becomes more spontaneous.

$$\Delta G$$ increases and the reaction becomes less spontaneous.

$$\Delta G$$ decreases and the reaction becomes more spontaneous.

$$\Delta G$$ decreases and the reaction becomes less spontaneous.

MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 7

A barometer made of very narrow tube is placed at normal temperature and pressure. The coefficient of volume expansion of mercury is

$0.00018$ per

$^oC$ and that of the tube is negligible. The temperature of mercury in the barometer is now raised by

$1^o$ C, but the temperature of the atmosphere does not change. Then the mercury height in the tube remains unchanged.

0%
True
0%
False

Explanation

Since volume of mercury increases the density of mercury decreases.

Non more height of the liquid will be req to lealance the atm pressure

$\boxed{B\ False}$

The specific heat of copper is

$0.385$ J/(g.

$^oC$ ). If

$34.2$ g of copper, initially at

$25^o$ C, absorbs

$4.689$ kJ, what will be the final temperature of the copper?

0%
$25.4^o$ C
0%
$27.8^o$ C
0%
$356^o$ C
0%
$381^o$ C

Explanation

$q=mc\Delta T$

Given that ,

$q=4689\;J,m=34.2,c=0.385\;Jg^{-}C^-$

Now,

$4689\; = (34.2)(0.385\;)(x - 25)$

$x = {381^ \circ }C$

If Q increases then:

0%
$\Delta G$ increases and the reaction becomes more spontaneous.
0%
$\Delta G$ increases and the reaction becomes less spontaneous.
0%
$\Delta G$ decreases and the reaction becomes more spontaneous.
0%
$\Delta G$ decreases and the reaction becomes less spontaneous.

Explanation

$Q = \Delta u+w$

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

considering work done

to be constant

$Q \uparrow \Delta u\uparrow$

$\Delta G \uparrow$

hence reactioon becomes spontaneous.

1170586_1033886_ans_ebad4706d99e4d378968973897bb7887.jpeg

Which heat is produced throughout the conducting wire?

0%
Petlier heat
0%
Thomson effect heat
0%
Joule heat
0%
none of these

Magnitude of Seebeck emf between the junctions does not depend on

0%
thermocouple
0%
temperature of cold junction
0%
temperature of hot function
0%
neutral temperature

The value of

$\triangle$ H and

$\triangle$ S for the reaction,

${ C }_{ \left( graphite \right) }+{ CO }_{ 2\left( g \right) }\rightarrow 2{ CO }_{ (g) }$ are 170kJ and

${ 170JK }^{ -1 }$ , respectively. This reaction will be spontaneous at:

0%
510K
0%
710K
0%
910K
0%
1119K

Explanation

$C_{(graphite)} + CO_{2}(g) \rightarrow 2CO(g)$

The

$Rn$ is spontaneous when

$\Delta G \leqslant O$

$\Rightarrow \Delta H - T\Delta S \leq O$

$\Rightarrow T\Delta S \geqslant \Delta H$

$\Rightarrow T\geqslant \dfrac{\Delta H}{\Delta S}$

$T \geqslant \dfrac{170 \times 10^{3}}{1170}$

$T \geqslant 1000K$

For an exothermic reaction to be sponteneous:

0%
temperature must be high
0%
temperature must be zero
0%
temperature may have any magnitude
0%
temperature and $\Delta H < T \Delta S$ .

Explanation

According to second law of thermodynamics,

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

For a reaction to be spontaneous,

$\Delta G$ have to be

$-ive$ .

For, an exothermic reaction,

$\Delta H$ is always

$-ve$ , which makes the value of

$\Delta G$

$-ve$ at any temperature.

So, exothermic reaction is spontaneous at any temperature.

So, correct answer is option C.

For a reversible spontaneous change

$\triangle s$ is:

0%
$\dfrac { \triangle E }{ T }$
0%
$\dfrac { P\triangle V }{ T }$
0%
$\dfrac { q }{ T }$
0%
$RTlogK$

Explanation

$ds=\dfrac{dq.rev}{T}$

$\displaystyle \triangle s =\dfrac{\int dq.rev}{T}$

$\displaystyle \int dq.rev.q$ (for rev. provess)

$\triangle s=\dfrac{q}{T}$

For a process to be spontaneous at constant

$T$ and

$P$ :

0%
$(\Delta G)_{system}$ must be negative
0%
$(\Delta G)_{system}$ must be positive
0%
$(\Delta S)_{system}$ must be positive
0%
$(\Delta S)_{system}$ must be negative

Explanation

The standard free energy change by using the standard entropies and enthalpies of a reaction.

When

$\Delta G=-ve$ , the change in the enthalpy and entropy will be positive and which is depends on the temperature and spontaneity ocuurs due to high temperature.

Which of the following must be spontaneous at all temperatures?

0%
$\Delta H_{sys}=-ve; \Delta S_{sys}=-ve$
0%
$\Delta H_{sys}=-ve; \Delta S_{sys}=+ve$
0%
$\Delta H_{sys}=+ve; \Delta S_{sys}=+ve$
0%
$\Delta H_{sys}=+ve:; \Delta S_{sys}=-ve$

Explanation

A reaction is spontaneous when the Gibbs free energy change (

$\Delta{G}<0$ )

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

where

$\Delta{H}$ is the change in enthalpy and

$\Delta{s}$ is the change in entropy.

$\Delta H=-Ve$

$\Delta S$ =

$-ve$ at low temperatures reaction is spontaneous.

$\Delta H=-Ve$

$\Delta S$ =

$+ve$ at all temperatures reaction is spontaneous.

$\Delta H=+Ve$

$\Delta S$ =

$+ve$ at high temperatures reaction is spontaneous.

$\Delta H=+Ve$

$\Delta S$ =

$-ve$ at all temperatures reaction is non-spontaneous.

Note: Low

$T$ and high

$T$ is relative to the entropy. Here, low

$T$ means the

$T-factor$ can not dominate

$\Delta{S}-factor$ . However, high

$T$ means that the

$T-factor$ will dominate

$\Delta{S}-factor$ .

Hence, the correct option is B.

Specific heats of monoatomic and diatomic gases are same and
satisfy the relation which is

0%
${C_p}\left( {mono} \right) = {C_p}\left( {dia} \right)$
0%
${C_p}\left( {mono} \right) = {C_\upsilon }\left( {dia} \right)$
0%
${C_\upsilon }\left( {mono} \right) = {C_\upsilon }\left( {dia} \right)$
0%
${C_\upsilon }\left( {mono} \right) = {C_p}\left( {dia} \right)$

$Pt$ |

$Cl_2$ (

$P_1$ atm) |

$HCl$ (

$0.1$ M) |

$Cl_2$ (

$P_2$ atm) |

$Pt$ , cell reaction will be spontaneous if:

0%
$P_1 = P_2$
0%
$P_1 > P_2$
0%
$P_2 > P_1$
0%
$P_1 = P_2 =1$ atm

Explanation

Solution:- (C)

${P}_{2} > {P}_{1}$

${E}_{cell} = {{E}^{0}}_{cell} + \cfrac{0.059}{2} \log{\cfrac{{P}_{2}}{{P}_{1}}}$

For cell reaction to be spontaneous,

${E}_{cell} > 0$

$\therefore {E}_{cell}$ will be

$+ve$ if

${P}_{2} > {P}_{1}$ .

Hence for the cell reaction to be spontaneous,

${P}_{2} > {P}_{1}$

In the series

$Sc(Z=21)$ to

$Zn(Z=30)$ , the enthalpy of atomisation of which element is least?

0%
Sc
0%
Mn
0%
Cu
0%
Zn

Explanation

$Sc\ \& \ Zn$ belongs to the third group of the periodic table. The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals, there are some unpaired electrons that account for their stronger metallic bonding.

Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in $Zn$ and as a result, it has the least enthalpy of atomization.

The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in

$ms^{-1}$ is : [

$k_B = 1.4 \times 10^{-23} \ J/K ; \ m_{He} = 7\times 10^{-27} \ kg$ ]

0%
$1.3 \times 10^4$
0%
$1.3 \times 10^3$
0%
$1.3 \times 10^5$
0%
$1.3 \times 10^2$

Explanation

Given,

$T=300K$

$m=7\times 10^{-27}kg$

$k_B=1.4\times 10^{-23}J/K$

The thermal velocity of helium is given by

$v=\sqrt{\dfrac{3k_BT}{m}}$

$v=\sqrt{\dfrac{3\times 1.4\times 10^{-23}\times 300}{7\times 10^{-27}}}$

$v=1.3\times 10^3m/s$

The correct option is B.

For the reaction of one mole zinc dust with one mole sulphuric acid in a bomb calorimeter,

$\triangle U$ and w correspond to?

0%
$\triangle U<0,w=0$
0%
$\triangle U<0,w<0$
0%
$\triangle U>0,w=0$
0%
$\triangle U>0,w>0$

Explanation

$Zn(s)+H_{2}SO_{4}(g)\rightleftharpoons ZnSO_{4}(s)+H_{2}(g)$

$\Delta _{n_{gas}}=0$

$\therefore w=\Delta _{n_{gas}}RT=0$

$\Delta U=w+q$

Bomb calorimeter is commonly used to find the heat of combustion of organic substances which consists of a sealed combustion chamber, called a bomb. If a process is run in a sealed container then no expansion or compression is allowed, so w = 0 and ∆U = q.

$\therefore q<0$

$\therefore \Delta U<0$

Hence, the correct option is

$\text{A}$

The specific heat of a metal is

$0.26$ . The chloride of the metal (always monomer) has its molecular mass

$95$ . The volume of hydrogen gas that

$1.2g$ of the metal will evolve at

${0}^{o}C$ and

$1$ atm, if it is allowed to react with excess of an acid, is ?

0%
$2.24L$
0%
$1.12L$
0%
$0.56L$
0%
$5.61L$

Explanation

Given specific heat

$=0.26$

$\therefore$ Atomic mass

$=\cfrac{6.4}{0.26}=24.61g$

(By dulong petit law)

$\therefore$ The metal is

$Mg.$

Mass of its chloride

$=95g.$

$\therefore$ Mass of chlorine

$=95-24.61=70.39g$

$\therefore$ The compound is

$MgCl_2$

Equivalent mass of

$MgCl_2=24.61$

$\therefore$ We have eq. mass

$=\cfrac{w\times 111.2}{V_0}$ (Hydrogen displacement methode)

$\therefore V_0=\cfrac{1.2\times 11.2}{24.61}=0.546L$

For a gas having molar mass M, specific heat at constant pressure can be given as:

0%
$\dfrac { \gamma R }{ M\left( \gamma -1 \right) }$
0%
$\dfrac { \gamma }{ RM }$
0%
$\dfrac { M }{ R\left( \gamma -1 \right) }$
0%
$\dfrac { \gamma RM }{ \gamma +1 }$

Explanation

$\gamma =\dfrac{C_{P}}{C_{V}}$

$\dfrac{C_{P}}{C_{P}-R}=\gamma \Rightarrow x=\gamma x-\gamma R$

$\Rightarrow x=\dfrac{\gamma R}{(\gamma -1)}$

Heat constant

$=\dfrac{x}{H}$

$=\dfrac{\gamma R}{(\gamma -1)M}$

For

${ Zn }^{ 2+ }/Zn$ ,

${ E }^{ \circ }=-0.76\quad V$ ,
for

${ Ag }^{ + }/Ag\quad { E }^{ \circ }=\quad 0.799\quad V$ .
The correct statement is:

0%

Zn undergoes reduction and Ag is oxidized
0%
in the reaction Zn getting reduced, Ag getting oxidized is spontaneous
0%
Zn undergoes oxidation $Ag^+$ gets reduced
0%
no suitable answer

Explanation

For feasible reaction,

$E_{cell} > 0$

$E^o Ag^+ / Ag > E^o \, zn^{2+}/ zn$

$\therefore Ag^+$ gets reduced and

$zn^-$ get oxidised to

$zn^{2+}$

Suppose that a reaction has

$\Delta H=40\ kJ$ and

$\Delta S=50\ kJ/ K$ . At what temperature range will it change from spontaneous to non-spontaneous?

0%
$0.8\ K$ to $1\ K$
0%
$799\ K$ to $800\ K$
0%
$800\ K$ to $801\ K$
0%
$799\ K$ to $801\ K$

Explanation

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

$\Delta H=$ 40 kJ

$\Delta S=$ 50 kJ/K

$\Delta G$ <0, process is spontaneous

$\Delta G$ >0, process is non spontaneous

$\Delta G$ =0, process is in equilibrium

For T

$=0.8$ K

$\Delta G=40-{0.8}\times{50}=0$

$\Delta G$ =0

The process is in equilibrium

For T

$=1$ K

$\Delta G=40-{1}\times{50}=-10$

The process is in spontaneous

For the reaction;

${ FeCO }_{ 3(s) }\longrightarrow { Fe }O_{ (s) }+C{ O }_{ 2 }(g)$ ,

$\Delta H=82.8\ kJ$ at

${ 25 }^{ 0 }C$ . What is

$\Delta U$ at

${ 25 }^{ 0 }C$ ?

0%
$82.8 kJ$
0%
$80.32 kJ$
0%
$-2394.77 kJ$
0%
$85.28 kJ$

Explanation

Solution:- (B)

$80.32 \; kJ$

${FeC{O}_{3}}_{\left( s \right)} \longrightarrow {FeO}_{\left( s \right)} + {C{O}_{2}}_{\left( g \right)}$

From first law of thermodynamics-

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

$\Delta{H} = 82.8 \; KJ$

$T = 25 ℃ = \left( 25 + 273 \right) K = 298 K$

$R = 8.314 \; J \; {mol}^{-1} {K}^{-1} =8.314 \times {10}^{-3} \; kJ \; {mol}^{-1} {K}^{-1}$

$\Delta{n} = {n}_{p} - {n}_{r} = \left( 1 \right) - 0 = 1$

$\therefore \Delta{U} = 82.8 - \left( 1 \times 8.314 \times {10}^{-3} \times 298 \right)$

$\Rightarrow \Delta{U} = 82.8 - 2.48 = 80.32 \; kJ$

Hence for th given reaction,

$\Delta{U}$ at

$25 ℃$ is

$80.32 \; kJ$ .

For a reaction to occur spontaneously:

0%
$\Delta S$ must be negative
0%
$(-\Delta H+T \Delta S)$ must be positive
0%
$\Delta H+T\Delta S$ must be negative
0%
$\Delta H$ must be negative

For a first order reaction rate constant is

$1 \times {10^{ - 5}}{\sec ^{ - 1}}$ having

${E_a} = 1800\ kJ/mol$ . Then the value of

$\ell nA$ at

$T = 600\ K$ is:-

0%
$151.7$
0%
$349.3$
0%
$24.7$
0%
$11.34$

Explanation

acc. to arrhenius equation

$k=A{ e }^{ -Ea/RT }\quad$

rate constant

$1\times { 10 }^{ -5 }{ sec }^{ -1 }$

$Ea=1800kJ/mol$

$\log { k } =\log { A } -\cfrac { Ea }{ RT }$

$\log { \left( 1\times { 10 }^{ -5 } \right) } =\log { A } -\cfrac { Ea }{ RT }$

$\log { A } =\log { \left( 1\times { 10 }^{ -5 } \right) } +\cfrac { 1800 }{ 1\times { 10 }^{ -5 }\times 600 } =151.7$

$\Delta S$ for

$4Fe_(s)+3O_{2(g)}\rightarrow 2Fe_2O_{3(s)}$ is

$550J/mol/K$ . The process is found to be spontaneous even at

$298K$

$[\Delta H=-1650kJ]$

0%
$\Delta S_{total}=-2000J$
0%
$\Delta S_{total}=+1650J$
0%
$\Delta S_{total }=+4980J$
0%
$\Delta S_total =-4980J$

$0.1 \,{m^3}$ of water at

${80^ \circ }$ is mixed with

$0.3 {m^3}$ of water at

${60^ \circ }$ . The final temperature of mixture is :

0%
${80^ \circ }C$
0%
${70^ \circ }C$
0%
${60^ \circ }C$
0%
${75^ \circ }C$

Explanation

${ m }_{ 1 }{ s }_{ 1 }{ dt }_{ 1 }={ m }_{ 2 }{ s }_{ 2 }{ dt }_{ 2 }$

$P\left[ 0.1 \right] \times 1\times \left[ 80-t \right] =P\left[ 0.3 \right] \times \left[ 1 \right] \times \left[ t-60 \right]$

$80-t=3t-180$

$4t=260=t=6{ 5 }^{ 0 }c$

The spontaneous nature of a reaction is impossible only if:

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

Using the relation,

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

$\triangle G$ is positive at all conditions of T and P only when

$\triangle H$ is positive and

$\triangle S$ is negative.

Since

$\triangle G$ is positive. The reaction is non-spontaneous which is not possible.

So, the correct option is

$D$

Which of the following set of condition necessarily implies a spontaneous process?

0%
Exothermic reaction with increasing disorder
0%
Endothermic reaction with increasing disorder
0%
Exothermic reaction with decreasing disorder
0%
Endothermic reaction with decreasing disorder

Explanation

Solution:- (A) Exothermic reaction with increasing disorder

For a spontaneous reaction

$\Delta{H} = -ve$

$\Delta{S} = +ve$

Spontaneity depends upon both critical minimum energy and maximum randomness or disorder.

Which of the natural process is spontaneous?

0%
Formation of curd from milk after doing initiation
0%
Convesion of $C (graphite) \rightarrow C (diamond)$ at $25^0$ C and $1$ atm
0%
Formation of $H_2$ (g) and $O_2$ (g) from $H_2O$ (l)
0%
Formation of $CrO_5$ from $H_2O_2$ and $K_2Cr_2O_7$ in basic medium

Explanation

(A) The curd forms from milk because of the chemical reaction between the lactic acid, bacteria and casein. During fermentation, the bacteria uses enzymes to produce energy

$(ATP)$ from lactose. This is spontaneous.

(B)

$C(graphite)\rightarrow C(diamond)$ at

$25^oC$ and

$1 atm$ . It is not spontaneous. It needs a

$150000$ times positive atmospheric pressure at Earth's surface to do so.

${H_2}_{(g)}$ and

${O_2}_{(g)}$ to

${H_2O}_{(l)}$ .

(D) It is not sponatneous. Acidified solution of

$K_2Cr_2O_7$ reacts with

$H_2O_2$ to form $$CrO_5$.

$Cr_2O_7^{2-}+2H^++H_2O_2 \longrightarrow CrO_5+5H_2O$

The specific heat of a metal is 0.16 cal/g approximate atomic weight would be:

0%
$40$
0%
$16$
0%
$32$
0%
$64$

Explanation

Solution:- (A)

$40$

According to Dulong's Petit law-

Approx. at. wt.

$\times$ Specific heat

$= 6.4$

$\therefore$ Approximate atomic weight of the metal

$= \cfrac{6.4}{0.16} = 40 \; g$ .

Hence the approximate atomic weight of the metal will be

$40 \; g$ .

Choose the correct statement.

0%
All spontaneous natural processes are thermodynamically irreversible.
0%
Work done is a state function.
0%
Work done is zero during an adiabatic expansion.
0%
For a spontaneous process it is always true that $\Delta S_{system}>0$ .

The incorrect criterion for the spontaneity of a process is:

0%
$\Delta S_{system }<0$
0%
$\Delta S_{system }+\Delta S_{surrounding }>0$
0%
$\Delta S_{system }>\Delta S_{surrounding }$
0%
$\Delta H_{system }<0$ and $\Delta S_{system}>0$

Explanation

For spontaneity of a process

${ \Delta S }_{ system }+{ \Delta S }_{ sourroundings }>0$

${ \Delta S }_{ system }>0$

${ \Delta S }_{ surroundings }<0$

Ans :- Option A.

Consider the following sequence of reactions
Which of the following is always true?

0%
$\Delta H_1=\Delta H_2+\Delta H_3+\Delta H_4$
0%
$\Delta H_2=\Delta H_1+\Delta H_3+\Delta H_4$
0%
$\Delta H_3=\Delta H_1+\Delta H_2+\Delta H_4$
0%
$\Delta H_4=\Delta H_1+\Delta H_2+\Delta H_3$

Explanation

Solution:- (B)

$\Delta{{H}_{2}} = \Delta{{H}_{1}} + \Delta{{H}_{3}} + \Delta{{H}_{4}}$

According to Hess's Law, regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

Hence,

$\Delta{{H}_{2}} = \Delta{{H}_{1}} + \Delta{{H}_{3}} + \Delta{{H}_{4}}$

An imaginary reaction

$X\rightarrow Y$ takes place in three steps

$X\rightarrow A,\Delta H=-q_{1}$ ;

$B\rightarrow A,\Delta H=-q_{2}$ ;

$Y\rightarrow B,\Delta H=-q_{3}$ If Hess' law is applicable, then the heat of the reaction

$X \rightarrow Y$ is:

0%
$q_{1}-q_{2}+q_{3}$
0%
$q_{2}-q_{1}+q_{3}$
0%
$q_{1}-q_{2}-q_{3}$
0%
$q_{3}-q_{2}-q_{1}$

Explanation

Solution:-

$X \longrightarrow A \quad \Delta{H} = - {q}_{1} ..... \left( 1 \right)$

$B \longrightarrow A \quad \Delta{H} = - {q}_{2}$

$A \longrightarrow B \quad \Delta{H} = {q}_{2} ..... \left( 2 \right)$

$Y \longrightarrow B \quad \Delta{H} = - {q}_{3}$

$B \longrightarrow Y \quad \Delta{H} = {q}_{3} ..... \left( 3 \right)$

Hess's Law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

Therefore,

Adding

${eq}^{n} \left( 1 \right), \left( 2 \right) \& \left( 3 \right)$ , we have

$X + A + B \longrightarrow A + B + Y \quad \Delta{H} = \left( -{q}_{1} \right) + \left( {q}_{2} \right) + \left( {q}_{3} \right)$

$X \longrightarrow Y \quad \Delta{H} = {q}_{2} - {q}_{1} + {q}_{3}$

Hence the heat of the reaction

$X \rightarrow Y$ is

${q}_{2} - {q}_{1} + {q}_{3}$ .

Which of the following is true?

0%
$C_{sp}=C_m \times$ molar mass
0%
$C_m=C_{sp} \times$ molar mass
0%
$C_m-C_{sp}=$ molar mass
0%
$C_m+C_{sp}=$ molar mass

Explanation

Solution:- (B)

${C}_{m} = {C}_{sp} \times \text{molar mass}$

Molar heat capacity

$\left( {C}_{m} \right)$ is a measure of the amount of heat necessary to raise the temperature of one mole of a pure substance by

$1 \; K$ .

Specific heat capacity

$\left( {C}_{sp} \right)$ is a measure of the amount of heat necessary to raise the temperature of one gram of a pure substance by

$1 \; K$ .

From the above definition, we can say that,

${C}_{m} = {C}_{sp} \times \text{molar mass}$

For a gaseous reaction involving the complete combustion of isobutane:

0%
$\triangle H = \triangle E$
0%
$\triangle H > \triangle E$
0%
$\triangle H < \triangle E$
0%
none of these

Explanation

Solution:- (C)

$\Delta{H} < \Delta{E}$

Combustion of isobutane-

${{C}_{4}{H}_{10}}_{\left( g \right)} + \cfrac{13}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow 4 {C{O}_{2}}_{\left( g \right)} + 5 {{H}_{2}O}_{\left( l \right)}$

For the above reaction-

$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = 4 - \left( \cfrac{13}{2} + 1 \right) = \cfrac{-7}{2} = -ve$

As we know that,

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

$\because \Delta{{n}_{g}} < 0$

$\therefore \Delta{H} < \Delta{E}$

$C(s) + O_2(g) \to CO_2(g); \quad \Delta H = -94kcal$

$2CO(g) + O_2(g) \to 2CO_2(g); \quad \Delta H = -135.2 kcal$

The heat of formation of

$CO(g)$ is :

0%
$-26.4\ kcal$
0%
$41.2\ kcal$
0%
$26.4\ kcal$
0%
$229.4\ kcal$

Explanation

Given,

$C(s) + O_2(g) \to CO_2; \Delta H_1 = -94Kcal.$ ....(1)

$2CO(g) + O_2(g) \to 2CO_2(g); \Delta H_2= - 135.2 K \ cal.$ ....(2)

$\Delta H_f(CO)=?$

Now the formation of

$CO$ is

$CO_2(g) + O_2(g) \to CO + \dfrac{1}{2}O_2(g)$ ....(3)

Then,

$\Delta H_3 = \dfrac{135.2}{2} = 67.6K\ cal$

$\therefore$ The required equation,

$C(s) + \dfrac{1}{2} O_2= CO(g); \Delta H_4 = ?$

Adding equation (1) and (3) we get

$C(s) + \dfrac{1}{2} O_2 \to CO(g)$ ;

$\Rightarrow \Delta H = -94+67.6 = -26.4K cal$ .

$\therefore$ The heat of formation of

$CO(g)$ is

$-26.4\ K\ cal$ .

Hence, te correct option is

$\text{A}$

Select the incorrect statement about the specific heats of a gaseous system?

0%
specific heat at no exchange condition, $C_{ A }=0$
0%
Specific heat at contant temperature, $C_{ T }=\infty$
0%
Specific heat at constant pressure, $C_{ P }=\frac { \gamma R }{ \gamma -1 }$
0%
Specific heat at constant volume, $C_{ V }=\frac { R }{ \gamma }$

Consider the reaction,

$2A_{(g)} + B_(g) rightleftarrow 2C_{(g)}$ for which

$K_c = 350$ . If

$0.01$ mole of each of the reactants and product are mixed in a

$2.0$ L flask, then the reaction quotient and the spontaneous direction of the system will be?

0%
$Q_c = 0.002$ ; the equilibrium shifts to the left
0%
$Q_c = 2000$ ; the equilibrium shifts to the left
0%
$Q_c = 0.002$ ; the equilibrium shifts to the right
0%
$Q_c = 2000$ ; the equilibrium shifts to the right

Consider the following reaction,

$2A + B \rightarrow C + 2D$ ,

$\Delta H_{1} = 10$

$A + 2C \rightarrow 2D + B$ ,

$\Delta H_{2} = -5$ What is

$\Delta H$ of reaction

$A + 2B \rightarrow 3C$ ?

0%
$-5$
0%
$+5$
0%
$+10$
0%
$+15$

Explanation

Solution:- (D)

$+15$

$A + 2B \longrightarrow 3C \quad \Delta{H} = ?$

Given:-

$2A + B \longrightarrow C + 2 D \quad \Delta{{H}_{1}} = 10 ..... \left( 1 \right)$

$A + 2C \longrightarrow 2D + B \quad \Delta{{H}_{2}} = -5$

$B + 2D \longrightarrow A + 2C \quad \Delta{{H}_{3}} = - \Delta{{H}_{2}} = 5 ..... \left( 2 \right)$

Adding

${eq}^{n} \left( 1 \right) \& \left( 2 \right)$ , we have

$2A + B + B + 2D \longrightarrow C + 2D + A + 2C \quad \Delta{H} = 10 + 5$

$A + 2B \longrightarrow 3C \quad \Delta{H} = 15$

Hence the

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

$+15$ .

For which one of the following system

$\Delta E < \Delta H$ ?

0%
$2SO_{2(g)}+O_{2(g)} \rightarrow 2SO_{3(g)}$
0%
$N_{2(g)}+O_{2(g)} \rightarrow 2NO_{(g)}$
0%
$2NH_{3(g)} \rightarrow N_{2(g)}+3H_{2(g)}$
0%
$H_{2(g)}+I_{2(g)} \rightarrow 2HI_{(g)}$

Explanation

Solution:- (C)

$2 {N{H}_{3}}_{\left( g \right)} \longrightarrow {{N}_{2}}_{\left( g \right)} + 3 {{H}_{2}}_{\left( g \right)}$

As we know that,

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

For

$\Delta{{n}_{g}} > 0$

$\Delta{H} > \Delta{E}$

Now for the reaction-

$2 {N{H}_{3}}_{\left( g \right)} \longrightarrow {{N}_{2}}_{\left( g \right)} + 3 {{H}_{2}}_{\left( g \right)}$

$\Delta{{n}_{g}} = \left( 3 + 1 \right) - 2 = +2 > 0$

$\therefore \Delta{H} > \Delta{E}$

Hence for the reaction

$2 {N{H}_{3}}_{\left( g \right)} \longrightarrow {{N}_{2}}_{\left( g \right)} + 3 {{H}_{2}}_{\left( g \right)}; \; \Delta{E} < \Delta{H}$ .

$300 \,K$ , the reactions which have following values of thermodynamic occurs spontaneously

0%
$\Delta G^o = -400 \,kJ \,mol^{-1}$
0%
$\Delta H^o = -200 \,kJ \,mol^{-1}, \Delta S^o = 4 \,kJK^{-1}mol^{-1}$
0%
$\Delta H^o = 200 \,kJ \,mol^{-1}, \Delta S^o = 4k \,JK^{-1} mol^{-1}$
0%
all of these

Explanation

For spontaneous reaction,

Gibbs free energy,

$\Delta{G}<0$

$\Delta S = +ve$

$\Delta G = -ve$

$\Delta H = -ve$

(i)

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

$\Delta{G}<O$

(ii)

$\Delta G = -200 - 300 \times 4=-1400$

$\Delta G = -ve$

(iii)

$\Delta G = 200 - 300 \times 4$

$\Delta G = -1000 - ve$

All the given options suggest a spontaneous reaction.

Hence, the correct option is D.

The pressure at the base of a column of liquid of length / and held at an angle

$\theta$ to the vertical is

0%
$\rho gl$
0%
$\rho gl \sin { \theta }$
0%
$\rho g\cos { \theta }$
0%
$\rho gl\cos { \theta }$

Explanation

Solutiion:- (D)

$\rho g l \cos{\theta}$

The pressure at the base of a column of liquid of length

$l$ and held at an angle

$\theta$ to the vertical is given as-

$P = \rho g l \cos{\theta}$

Which of the following expression for an irreversible process:

0%
$d{S}_{Total}> \cfrac{dq}{T}$
0%
$d{S}_{Total}= \cfrac{dq}{T}$
0%
$d{S}_{Total}< \cfrac{dq}{T}$
0%
$d{S}_{Total}= \cfrac{dU}{T}$

Calculate the amount of heat necessary to raise

$213.5 g$ of water from

${25}^{o}$ and

${100}^{o}C$ Molar heat capacity of water is

$18 cal$

${mole}^{-1}{K}^{-1}$

0%
$16.0125 kcal$
0%
$12.0125 kcal$
0%
$20.222 kcal$
0%
$26.323 kcal$

Identify the correct statement regarding a spontaneous process.

0%
For a spontaneous process in an isolated system, the change in entropy is positive
0%
Endothermic processes are never spontaneous
0%
Exothermic processes are always spontaneous
0%
Lowering of energy in the reaction process is the only criterion for spontaneity

Heat of formation of

$H_2O$ (g) at

$25^0C$ is -243 KJ.

$\Delta U$ for the reaction

$H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O (g)$ at

$25^0C$ is:

0%
241.8 KJ
0%
-241.8 KJ
0%
-243 KJ
0%
243 KJ

Explanation

Solution:- (B)

$-241.8 \; kJ$

${{H}_{2}}_{\left( g \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {{H}_{2}O}_{\left( g \right)} \quad \Delta{H} = -243 \; kJ$

For the above reaction-

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

Given:-

$T = 25 ℃ = \left( 25 + 273 \right) = 298 K$

As we know that,

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

$\Rightarrow -243 = \Delta{U} + \left( \cfrac{-1}{2} \times 8.314 \times {10}^{-3} \times 298 \right)$

$\Rightarrow \Delta{U} = -243 + 1.2 = - 241.8 \; kJ$

Hence the value of

$\Delta{U}$ for the given reaction is

$-241.8 \; kJ$ .

Suppose that a reaction has

$\Delta H=-40kJ$ and

$\Delta S=-50J/K$ . At what temperature range will it change from spontaneous to non-spontaneous ?

0%
$0.79\ K$ to $0.81\ K$
0%
$799\ K$ to $801\ K$
0%
$801\ K$ to $799\ K$
0%
$0.81\ K$ to $0.79\ K$

$\Delta H$ and

$\Delta S$ are

$-283\ kJ$ and

$-87\ J\ K^{-1}$ , respectively. It was intended to carry out this reaction at

$1000, 1500, 3000$ and

$3500\ K$ . At which of these temperatures would this reaction be thermodynamically spontaneous?

0%
$1000, 1500$ and $2000\ K$
0%
$500, 3000$ and $3500\ K$
0%
$1500$ and $3500\ K$
0%
$3000$ and $3500\ K$

Which of the following statements is wrong?

0%
Evaporation is a spontaneous process.
0%
Evaporation is a surface phenomenon.
0%
Vapour pressure decreases with increase of temperature.
0%
The vapour pressure of a solution is always less than the vapour pressure of a pure solvent.

Which of the following is not true about a reversible reaction?

0%
The reaction does not proceed to completion
0%
It cannot be influenced by a catalyst
0%
Number of moles of reactants and products is always equal
0%
It can be attained only in a closed container

For the reaction:

${ C }_{ 2 }{ H }_{ 4 }(g)+3{ O }_{ 2 }(g)\rightarrow 2C{ O }_{ 2 }(g)+2{ H }_{ 2 }O(l)$

$\Delta E=-1415KJ$ . The

$\Delta H$ at

$27C$ is:

0%
$-1410 KJ$
0%
$-1420 KJ$
0%
$+1420 KJ$
0%
$+1410 KJ$

Explanation

Solution:- (B)

$-1420 \; kJ$

Given:-

$\Delta{E} = - 1415 \; kJ$

$T = 27 ℃ = \left( 27 + 273 \right) = 200 \; K$

For the given reaction-

${{C}_{2}{H}_{4}}_{\left( g \right)} + 3 {{O}_{2}}_{\left( g \right)} \longrightarrow 2 {C{O}_{2}}_{\left( g \right)} + 2 {{H}_{2}O}_{\left( l \right)}$

$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = 2 - \left( 3 + 1 \right) = -2$

Now as we know that,

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

$\Rightarrow \Delta{H} = -1415 + \left( -2 \times 8.314 \times {10}^{-3} \times 300 \right)$

$\Rightarrow \Delta{H} = - 1415 - 5 = -1420 \; kJ$

Hence the value of

$\Delta{H}$ is

$-1420 \; kJ$ .

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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