MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 14

When two mole of an ideal gas

(C_{p . m} = \frac{5}{2} R)

$\left( { C }_{ p.m }=\cfrac { 5 }{ 2 } R \right)$ heated from

300 K

$300K$ to

600 K

$600K$ at constant pressure. The change in entropy of gas (

Δ S

$\Delta S$ ) is:

0%
$-\cfrac{3}{2}R\ln{2}$
0%
$\cfrac{3}{2}R\ln{2}$
0%
$5R\ln {2}$
0%
$\cfrac{5}{2}R\ln{2}$

Explanation

Solution:- (C)

5 R ln 2

$5R \ln{2}$

Δ S = n C_{P} ln \frac{T_{2}}{T_{1}} + n R ln \frac{P_{1}}{P_{2}}

$\Delta{S} = n{C}_{P} \ln{\cfrac{{T}_{2}}{{T}_{1}}} + nR \ln\dfrac{{P}_{1}}{{P}_{2}}$

$\because$ Pressure is constant,

$\Delta{S} = n {C}_{P} \ln{\cfrac{{T}_{2}}{{T}_{1}}}$

Given:-

$n = 2 \text{ moles}$

${C}_{P} = \cfrac{5}{2}R$

${T}_{2} = 600 K$

${T}_{1} = 300 K$

$\therefore \Delta{S} = 2 \times \cfrac{5}{2} R \times \ln{\cfrac{600}{300}}$

$\Rightarrow \Delta{S} = 5R \ln{2}$

Hence the change in entropy of the gas is

$5R \ln{2}$ .

For the reaction

$A\rightleftharpoons B+C$ . At equilibrium, the concentration of

$A$ is

$1\times { 10 }^{ -3 }M$ ,

$B$ is

$0.15M$ and

$C$ is

$0.05M$ . The

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

$A$ at

${ 27 }^{ o }C$ will be

0%
$-5kJ/mol$
0%
$-17.3kJ/mol$
0%
$-11.2kJ/mol$
0%
$none\ of\ these$

Which of the following is/are correct ?

0%
$\Delta H=\Delta U+\Delta$ (PV) when P and V both changes
0%
$\Delta H=\Delta U+P\Delta V$ when pressure is constant
0%
$\Delta H=\Delta U+V\Delta P$ when volume is constant
0%
all are correct

Which of the following processes are spontaneous

0%
Melting of ice at $2$ atm and $273K$
0%
Melting point of ice at $1/2$ atm and $273K$
0%
Boiling of water at $1/2$ atm and $373K$
0%
Boiling of water at $2$ atm and $373K$

For a reaction.

$A(g)\rightarrow A(l);\Delta {H}=-3RT$ . The correct statement for the reaction is:

0%
$\left| \Delta H \right| >\left| \Delta U \right|$
0%
$\Delta H=\Delta U\ne 0$
0%
$\left| \Delta H \right| <\left| \Delta U \right|$
0%
$\Delta H=\Delta U= 0$

Explanation

Solution:- (A)

$\left| \Delta{H} \right| > \left| \Delta{U} \right|$

${A}_{\left( g \right)} \longrightarrow {A}_{\left( l \right)}$

For the above reaction,

$\Delta{{n}_{g}} = 0 - 1 = -1$

Now as we know that,

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

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

$\Rightarrow \Delta{U} = -3RT - \left( -1 \right) RT = -2 RT$

$\left| \Delta{H} \right| = \left| -3RT \right| = 3RT$

$\left| \Delta{U} \right| = \left| -2RT \right| = 2RT$

Hence

$\left| \Delta{H} \right| > \left| \Delta{U} \right|$

For the spontaneous process

$2F(g)\rightarrow {F}_{2}(g)$ , the sign of

$\Delta H$ and

$\Delta S$ respectively are?

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

The favourable conditions for a spontaneous reaction are:

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

Explanation

The free energy charge

$\triangle G$ is given by,

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

The reaction is spontaneous if

$\triangle G<0.$

$\Longrightarrow \triangle H-T\triangle S<0$

$\Longrightarrow T\triangle S>\triangle H, \triangle H=+ve,\triangle S=+ve$

For a reaction

${R}_{1},\Delta G=x KJ {mol}^{-1}$ . For a reaction

${R}_{2},\Delta G=y$

$KJ{mol}^{-1}$ . Reaction

${R}_{1}$ is non spontaneous but along with

${R}_{2}$ it is spontaneous. This means that:

0%
$x$ is $-ve$ , $y$ is $+ve$ but in magnitude $x> y$
0%
$x$ is $-ve$ , $y$ is $+ve$ but in magnitude $y> x$
0%
Both $x$ and $y$ are $-ve$ but not equal
0%
Both $x$ and $y$ are $+ve$ but not equal

Which of the following conditions make the process non spontaneous at all temperatures.

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

Consider the formation of

$MgO(s)$ . Assume that

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

$\triangle { S }^{ o }$ are independent of temperature:

$Mg(s)+\cfrac { 1 }{ 2 } { O }_{ 2 }\left( g \right) \rightarrow MgO\left( s \right)$

$\triangle { H }^{ o }=-602kJ/mol$ ,

$\triangle { S }^{ o }=-108kJ/mol$

Calculate

$\triangle { G }$ for formation of

$MgO(s)$ at

${0}^{o}C$ and is the reacion spontaneous or non spontaneous at

${0}^{o}C$ ?

0%
$29,000 kJ/mol,$ spontaneous
0%
$-29,000 kJ/mol,$ non spontaneous
0%
$-30,000 kJ/mol,$ spontaneous
0%
$30,000 kJ/mol,$ non spontaneous

For a spontaneous reaction, the

$\Delta G$ , equilibrium constant,

$K$ and will be respectively:

0%
$-ve,> 1$
0%
$+ve, 1$
0%
$-ve,< 1$
0%
$-ve. 1$

For a gaseous reaction,

$A(g)+3B(g)\rightarrow 3C(g)+3D(g)$

$\Delta E$ is

$17kcal$ at

${27}^{o}C$ , assuming

$R=2cal$

${K}^{-1}$

${mol}^{-1}$ , the value of

$\Delta H$ for the above reaction is:

0%
$15.8kcal$
0%
$18.2kcal$
0%
$20.0kcal$
0%
$16.4kcal$

$\Delta {G}^{o}$ of

${Cu}_{(aq)}^{+}$ and

${Cu}_{(aq)}^{2+}$ respectively are

$+50$ and

$+66kJ/mole$ . Value of (

$\Delta{H}^{o}-T\Delta {S}^{o}$ ) for

${Cu}_{(aq)}^{+}\rightarrow {Cu}_{(aq)}^{2+}$ in kilo joules is?

0%
$-16$
0%
$+116$
0%
$-116$
0%
$+16$

Which plot represents for an exothermic reaction:

Explanation

For exothermic reaction:-

$\Delta H= negative$

$\Delta H= H_P- H_R$

$= negative$

So,

$H_R > H_P$ (for exothermic reaction)

(Refer to Image)

This plot represents exothermic reaction.

1094535_1144935_ans_d5b47702b3b74039a28dd4f35c2c2dc0.png

Consider the reaction:

$4 NO_{2(g)} + O_{2(g)} \rightarrow 2 N_2O_{5(g)} ; \, \, \, \Delta_rH = -111 kJ$

$N_2O_{5(s)}$ is formed instead of

$N_2O_{5(g)}$ in the above reaction, then

$\Delta_rH$ value will be:

[Given,

$\Delta H$ of sublimation for

$N_2O_5$ is

$54 \, kJ \, mol^{-1}$ ]

0%
$+ 54\ kJ$
0%
$+ 219\ kJ$
0%
$-219\ kJ$
0%
$-165\ kJ$

Explanation

$4 {N{O}_{2}}_{\left( g \right)} + {{O}_{2}}_{\left( g \right)} \longrightarrow 2 {{N}_{2}{O}_{5}}_{\left( g \right)}; \quad \Delta{H} = -111 \; kJ ..... \left( 1 \right)$

${{N}_{2}{O}_{5}}_{\left( s \right)} \longrightarrow {{N}_{2}{O}_{5}}_{\left( g \right)}; \quad \Delta{H} = 54 \; kJ$

${{N}_{2}{O}_{5}}_{\left( g \right)} \longrightarrow {{N}_{2}{O}_{5}}_{\left( s \right)}; \quad \Delta{H} = -54 \; kJ$

$2 \times \left[ {{N}_{2}{O}_{5}}_{\left( g \right)} \longrightarrow {{N}_{2}{O}_{5}}_{\left( s \right)}; \quad \Delta{H} = -54 \; kJ \right]$

$2 {{N}_{2}{O}_{5}}_{\left( g \right)} \longrightarrow 2 {{N}_{2}{O}_{5}}_{\left( s \right)}; \quad \Delta{H} = -108 \; kJ ..... \left( 2 \right)$

Adding

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

$4 {N{O}_{2}}_{\left( g \right)} + {{O}_{2}}_{\left( g \right)} + 2 {{N}_{2}{O}_{5}}_{\left( g \right)} \longrightarrow 2 {{N}_{2}{O}_{5}}_{\left( g \right)} + 2 {{N}_{2}{O}_{5}}_{\left( s \right)}; \quad \Delta{H} = \left[ \left( -111 \right) + \left( -108 \right) \right] \; kJ$

$4 {N{O}_{2}}_{\left( g \right)} + {{O}_{2}}_{\left( g \right)} \longrightarrow 2 {{N}_{2}{O}_{5}}_{\left( s \right)}; \quad \Delta{H} = -219 \; kJ$

Hence if

${{N}_{2}{O}_{5}}_{\left( s \right)}$ is formed instead of

${{N}_{2}{O}_{5}}_{\left( g \right)}$ , the value of

$\Delta{H}$ will be

$-219 \; kJ$ .

$\Delta{S}_{surroundings}=+959.1J{K}^{-1}{mol}^{-1}$

$\Delta{S}_{system}=-163.1J{K}^{-1}{mol}^{-1}$ . Then process is?

0%
Spontaneous
0%
Non spontaneous
0%
At equilibrium
0%
Cannot be predicted from the information

$\Delta S$ for

$4Fe(s)+3{O}_{2}(s)\rightarrow 2{Fe}_{2}{O}_{3}(s)$ is

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

$298K$ because (

$\Delta H=-1650kJ$ )

0%
$\Delta {S}_{total}=-2000J$
0%
$\Delta {S}_{total}=+1650J$
0%
$\Delta {S}_{total}=+4980J$
0%
$\Delta {S}_{total}=-4980J$

In which of the following case entropy decreases-

0%
Solid changing to liquid
0%
Expansion of a gas
0%
Crystals dissolve
0%
Polymerisation

Ethyl chloride (

${C}_{2}{H}_{5}Cl$ ) is prepared by reaction of ethylene with hydrogen cloride:

${C}_{2}{H}_{4}(g)+Hcl(g)\rightarrow {C}_{2}{H}_{5}Cl(g)$ ;

$\Delta H=-72.3kJ$
What is the value of

$\Delta E$ (in kJ), if

$70g$ of ethylene and

$73g$ of

$HCl$ are allowed to react at

$300K$

0%
$-69.8$
0%
$-180.75$
0%
$-174.5$
0%
$-139.6$

Consider the following processes

$\triangle H(KJ/mol)$

$\frac { 1 }{ 2 } A\rightarrow B$

$+150$

$3B\rightarrow 2C+D$

$-125$

$E+A\rightarrow D$

$+350$
For

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

$\triangle H$ will be :

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

An equilibrium reaction

$X+Y \rightleftharpoons W+Z, \triangle H=+ve$ is spontaneous in the forward direction. Then corresponding sign of

$\triangle G$ and

$\triangle S$ should be respectively:

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

Rank the following substances in order of decreasing heat of combusion (maximum

$\to$ minimum).

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

In an insulated container 1 mole of a liquid. molar volume 100 ml at 1 bar. Liquid is steeply taken to 100 bar, when volume of liquid decreases by 1 ml. Find

$\Delta H$ for the process.

0%
7900 bar mL
0%
8900 bar mL
0%
9900 bar mL
0%
10900 bar mL

the above reaction was carried out at 300 K in a bomb calorimeter. The heat released was 743 kJ/mol. The value of

$\triangle H_{300K}$ for this reaction would be:

0%
-740.5 kJ/mol
0%
-741.75 kJ/mol
0%
-743.0 kJ/mol
0%
-744.25 kJ/mol

For the reaction

$A\rightleftharpoons B+C$ At the equilibrium, the concentration of

$A$ is

$1\times { 10 }^{ -3 }M$ is

$B$ is

$0.15M$ and

$C$ is

$0.05M$ . The

$\triangle { G }^{ o }$ for the reaction at

${ 27 }^{ o }C$ will be

0%
$-5kJ/ mol$
0%
$-17.36kJ/ mol$
0%
$-11.2kJ/ mol$
0%
$10.1kJ/mol$

Given the following data:
Substance

$\Delta H$ (KJ/mol)

${S^ \circ }$ (J/mol K)

$\Delta G\,\,\,$ (KJ/mol)
FeO(s) -266.3 57.49 -245.12
C(Graphite) 0 5.74 0
Fe(s) 0 27.28 0
CO(g) -110.5 197.6 -137.15
Determine at what temperature the following reaction is spontaneous ?

${\text{FeO}}\left( {\text{s}} \right){\text{ + C}}\left( {{\text{Graphite}}} \right) \to {\text{Fe}}\left( {\text{s}} \right){\text{ + CO}}\left( {\text{g}} \right)$

0%
289 K
0%
668 K
0%
966 K
0%
${{\text{G}}^{\text{o}}}$ is ${\text{ + ve}}$ . hence the reaction will never be spontaneous

For the process,

${\text{C}}{{\text{O}}_{\text{2}}}\left( s \right) \to \,{\text{C}}{{\text{O}}_{\text{2}}}\left( g \right)$

0%
Both ${{\Delta H}}\,\,\,\,$ and ${{\Delta S}}$ are negative
0%
${{\Delta H}}\,\,\,\,$ is negative and ${{\Delta S}}$ is positive
0%
${{\Delta H}}\,\,\,\,$ is +Ve and ${{\Delta S}}$ is negative
0%
Both ${{\Delta H}}\,\,\,\,$ and ${{\Delta S}}$ are positive

Spontaneity may be observed in following conditions

0%
$\triangle H$ = -ve, $\triangle S$ = +ve
0%
$\triangle H$ = +ve, $\triangle S$ = +ve
0%
$\triangle H$ = -ve, $\triangle S$ = -ve
0%
All of these

Explanation

When the process is exothermic

$(\triangle H _{system} < 0 ),$

and the entropy of the system increases

(

$\triangle$ system

$> 0), \triangle G$ will be negative at

all temperatures. Thus, the process is always

spontaneous.

1188628_1180495_ans_7e9cc8fa138845718d52961d46d1d7cb.jpg

Using the data provided, calculate the multiple bond energy (kJ / mol) of a

$C=C$ bond in

$C_{2}H_{2}$ . (Take the bond energy of a C-H bond as 350 kJ / mol)

$2C(s) + H_{2}(g) \rightarrow C_{2}H_{2}(g)$

$\Delta H = 225 kJ / mol$

$2C(s)\rightarrow 2C(g);\Delta H= 1410kJ mol^{-1}$

$H_2(g)\rightarrow 2H(g);\Delta H= 330kJ mol^{-1}$

0%
1165
0%
837
0%
865
0%
815

In a reaction, the change in entropy is given as 2.4 cal/K and the change in Gibbs free energy is given as 3.4 kcal, calculate the change in heat at the temperature of 20-degree centigrade?

0%
3.4 kcal
0%
3.4 cal
0%
3.4 kJ
0%
3.4 J

For the reaction between

$CO_2$ and graphite

$CO_2 (g) + C(s) \rightarrow 2CO(g)$

$\Delta H = 170.0 KJ and \Delta s = 170 JK^{-1}$ . The reaction is spontaneous at

0%
298 K
0%
500 K
0%
900 K
0%
none of the above

An endothermic reaction is spontaneous if

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

For the reaction,

$A\rightarrow B,\Delta H=+ve,\Delta S=-ve$ This reaction is

0%
non-spontaneous at all temperature
0%
non-spontaneous at low temperature
0%
non-spontaneous at high temperature
0%
spontaneous at high temperature

For the reaction

$X_2Y_4(l)\rightarrow 2 XY_2(g)$ at 300 K the values of

$\Delta U$ and

$\Delta S$ are

$2 kcal$ and

$20 cal K^{-1}$ respectively. The value of

$\Delta G$ for the reaction is:

0%
-3400 cal
0%
3400 cal
0%
2000 cal
0%
-2800 cal

The reaction

$2A(g)\rightarrow A_{2}(g)$ , will be spontaneous

0%
At high temperature
0%
At low temperature
0%
At all temperature
0%
Never at any temperature

The correct thermodynamic conditions for the spontaneous reaction at all temperature is:

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

Consider the following processes:

$\frac { 1 }{ 2 } A\longrightarrow B:\quad \quad \quad \quad \quad \triangle H=150$

$3B\longrightarrow 2C+D:\quad \quad \quad \quad \quad \triangle H=-125$

$E+A\longrightarrow 2D\quad \quad \quad \quad \quad \quad \quad \triangle =350$
For B+D

$\rightarrow$ E+2C,

$\triangle$ H will be:

0%
325 kJ/mol
0%
525 kJ/mol
0%
-175 kJ/mol
0%
-325 kJ/mol

The difference between heat of reaction at pressure and constant volume for the reaction,

${ CH }_{ 2 }={ CH }_{ 2 }(g)+{ 3O }_{ 2 }(g)\rightarrow { 2CO }_{ 2 }(g)+{ 2H }_{ 2 }O(l)$ at 300 K is?

0%
5.87 kJ
0%
-4.99 kJ
0%
6.89 kJ
0%
-7.25 kJ

Given that

$Zn+1/2O_2\rightarrow ZnO+35.25kJ. HgO\rightarrow Hg+1/2O_2+9.11kJ.$ The heat of the reaction

$Zn+HgO\rightarrow ZnO+Hg$ is

0%
-26.14kJ
0%
44.39kJ
0%
-44.39kJ
0%
26.14kJ

Explanation

$Zn+\cfrac{1}{2}O_{2}\rightarrow \,ZnO+35.25\,kJ$

$HgO\rightarrow \,Hg+\cfrac{1}{2}O_{2}+9.11\,kJ$

By Adding both we get,

$Zn+HgO\rightarrow ZnO+Hg+44.36\,kJ$

Heat of reaction is

$-44.39\,\,kJ$

What is the free energy change

$\triangle G$ , when

$1.0$ mole of water at

$100^{o}C$ and

$1$ atm pressure is converted steam at

$100^{o}C$ and

$1$ atm pressure:-

0%
+540 cal
0%
-9800 cal
0%
+9800 cal
0%
0 cal

Explanation

As we know that

$\Delta G$ at equilibrium is always

$0$ . So, here also is equilibrium ; so

$\Delta G=0$

500 J of heat was supplied to a system at constant volume.It resulted in the increase of temperature of the system from

$20^oC$ to

$25^oC$ .What is the change in internal energy of the system?

0%
$+8.43\ J$
0%
$+43\ J$
0%
$+458.43\ J$
0%
$+580.43\ J$

The P - V diagram of a system undergoing a thermodynamic process is shown in figure. Work done by the system in going from

$A \rightarrow B \rightarrow C$ is 30 J and 40 J heat is given to the system. The change in internal energy of the gas if the gas is directly taken from A to C is

0%
10 J
0%
70 J
0%
84 J
0%
134 J

Explanation

For process

$A \to B \to C$

$\Delta u = \Delta \theta - \Delta \omega$

$= 40 - 30$

$= 10J$

$A \to C$

$\Delta u = 10J$

Hence

$,$ option

$(A)$ is correct

$.$

$2A{ l }_{ 2 }(s)\longrightarrow 4Al(s)+3{ O }_{ 2 }(g),\triangle G=+138$ kcal considering the contributing of entropy to the spontaneity of this reaction, the reaction is________And entropy of the system_______.

0%
Spontaneous increase
0%
Spontaneous decrease
0%
non-Spontaneous, increase
0%
non-Spontaneous decrease

For reaction

$A \rightarrow B$ ,

$\Delta H$ and

$\Delta S$ are positive. the most favourable condition of spontaneous process.

0%
low temperature
0%
high temperature
0%
high concentration
0%
very low temperature

The Gibb's energy for the decomposition of

$Al_{2}O_{3}$ at

$500^oC$ is as follows:

$2/3A{ l }_{ 2 }{ O }_{ 3 }\longrightarrow 4/3Al+{ O }_{ 2 }{ \triangle }_{ r }G=+966\quad kJ\quad { mol }^{ -1 }$
The potential difference needed for electrolytic reduction of

$Al_{2}O_{3}$ at

$500^oC$ is at least

0%
5.0 V
0%
4.5 V
0%
3.0 V
0%
2.5 V

Dextrorotatory

$\alpha$ -pinene has a specific rotation

$[ \alpha ] _ { D } ^ { 20 } = + 51.3 ^ { \circ } .$ A sample of

$\alpha -pinene$ containing both the enantiomers was found to have a specific rotation value

$[ \alpha ] _ { D } ^ { 20 } = + 30.8 ^ { \circ } .$ The percentages of the

$( + )$ and

$( - )$ enantiomers present in the sample are, respectively.

0%
$70 \%$ and $30 \%$
0%
$80 \%$ and $20 \%$
0%
$20 \%$ and $80 \%$
0%
$60 \%$ and $40 \%$

$\begin{array}{l}N{H_3}\left( g \right) + 3C{l_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} NC{l_4}\left( g \right) + 3HCl: - \Delta {H_1}\\{N_2}\left( g \right) + 3{H_2}\left( g \right) \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2N{H_3}\left( g \right):\Delta {H_2}\\{H_2}\left( g \right) + C{l_2}\left( g \right) \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2HCl\left( g \right):\Delta {H_3}\end{array}$
The heat of formation of

$NC{l_4}$ in the terms of

$\Delta {H_1},\Delta {H_2},\Delta {H_3}$ is:

0%
$\Delta {H_1} = - \Delta {H_1} + \dfrac{{\Delta {H_2}}}{2} - \dfrac{3}{2}\Delta {H_3}$
0%
$\Delta {H_1} = \Delta {H_1} + \dfrac{{\Delta {H_2}}}{2} - \dfrac{3}{2}\Delta {H_3}$
0%
$\Delta {H_1} = \Delta {H_1} - \dfrac{{\Delta {H_2}}}{2} + \dfrac{3}{2}\Delta {H_3}$
0%
none of these

An ideal gas expands according to the law

$P^{2}V$ = constant. The internal energy of the gas

0%
Increases continuously
0%
Decreases continuously
0%
Remain constant
0%
First increases and then decreases

In which of the following ionic compounds,

$\Delta H_f$ is negative only due to lattice energy:
(i)

$NaF$ (ii)

$MgO$ (iii)

$Li_2N$ (iv)

$Na_2S$

0%
Only (iv)
0%
Only (iii) and (iv)
0%
Only (ii), (iii), (iv)
0%
All of these

When a bottle of perfume is opened, odorous molecules mix with air and slowly diffuse throughout the entire room. The incorrect fact about the process is:

0%
$\triangle G = -Ve$
0%
$\triangle H \simeq 0$
0%
$\triangle S = -Ve$
0%
$\triangle S = +ve$

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 14 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 14 - MCQExams.com

0:0:3

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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