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CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 14 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Thermodynamics
Quiz 14
When two mole of an ideal gas $$\left( { C }_{ p.m }=\cfrac { 5 }{ 2 } R \right) $$ heated from $$300K$$ to $$600K$$ at constant pressure. The change in entropy of gas ($$\Delta S$$) is:
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$$-\cfrac{3}{2}R\ln{2}$$
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$$\cfrac{3}{2}R\ln{2}$$
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$$5R\ln {2}$$
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$$\cfrac{5}{2}R\ln{2}$$
Explanation
Solution:- (C) $$5R \ln{2}$$
$$\Delta{S} = n{C}_{P} \ln{\cfrac{{T}_{2}}{{T}_{1}}} + nR \ln\dfrac{{P}_{1}}{{P}_{2}}$$
$$\because$$ 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
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$$-5kJ/mol$$
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$$-17.3kJ/mol$$
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$$-11.2kJ/mol$$
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$$none\ of\ these$$
Which of the following is/are correct ?
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$$\Delta H=\Delta U+\Delta $$ (PV) when P and V both changes
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$$\Delta H=\Delta U+P\Delta V$$ when pressure is constant
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$$\Delta H=\Delta U+V\Delta P$$ when volume is constant
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all are correct
Which of the following processes are spontaneous
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Melting of ice at $$2$$ atm and $$273K$$
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Melting point of ice at $$1/2$$ atm and$$273K$$
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Boiling of water at $$1/2$$ atm and $$373K$$
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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:
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$$\left| \Delta H \right| >\left| \Delta U \right| $$
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$$\Delta H=\Delta U\ne 0$$
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$$\left| \Delta H \right| <\left| \Delta U \right| $$
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$$\Delta H=\Delta U= 0$$
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?
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+ve, -ve
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+ve, +ve
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-ve, -ve
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-ve, +ve
The favourable conditions for a spontaneous reaction are:
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$$T\Delta S> \Delta H, \Delta H=+ve,\Delta S=+ve$$
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$$T\Delta S> \Delta H, \Delta H=+ve,\Delta S=-ve$$
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$$T\Delta S= \Delta H, \Delta H=+ve,\Delta S=-ve$$
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$$T\Delta S= \Delta H, \Delta H=+ve,\Delta S=+ve$$
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:
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$$x$$ is $$-ve$$, $$y$$ is $$+ve$$ but in magnitude $$x> y$$
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$$x$$ is $$-ve$$, $$y$$ is $$+ve$$ but in magnitude $$y> x$$
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Both $$x$$ and $$y$$ are $$-ve$$ but not equal
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Both $$x$$ and $$y$$ are $$+ve$$ but not equal
Which of the following conditions make the process non spontaneous at all temperatures.
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$$\Delta H=+ve;\Delta S=-ve$$
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$$\Delta H=-ve;\Delta S=+ve$$
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$$\Delta H=+ve;\Delta S=+ve$$
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$$\Delta H=-ve;\Delta S=-ve$$
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 $${0}^{o}C$$?
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$$29,000 kJ/mol, $$ spontaneous
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$$-29,000 kJ/mol, $$ non spontaneous
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$$-30,000 kJ/mol, $$ spontaneous
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$$30,000 kJ/mol, $$ non spontaneous
For a spontaneous reaction, the $$\Delta G$$, equilibrium constant, $$K$$ and will be respectively:
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$$-ve,> 1$$
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$$+ve, 1$$
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$$-ve,< 1$$
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$$-ve. 1$$
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:
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$$15.8kcal$$
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$$18.2kcal$$
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$$20.0kcal$$
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$$16.4kcal$$
$$\Delta {G}^{o}$$ 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?
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$$-16$$
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$$+116$$
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$$-116$$
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$$+16$$
Which plot represents for an exothermic reaction:
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0%
0%
0%
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.
Consider the reaction:
$$4 NO_{2(g)} + O_{2(g)} \rightarrow 2 N_2O_{5(g)} ; \, \, \, \Delta_rH = -111 kJ$$
If $$N_2O_{5(s)}$$ 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 $$54 \, kJ \, mol^{-1}$$]
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$$+ 54\ kJ$$
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$$+ 219\ kJ$$
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$$-219\ kJ$$
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$$-165\ kJ$$
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?
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Spontaneous
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Non spontaneous
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At equilibrium
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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 ($$\Delta H=-1650kJ$$)
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$$\Delta {S}_{total}=-2000J$$
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$$\Delta {S}_{total}=+1650J$$
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$$\Delta {S}_{total}=+4980J$$
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$$\Delta {S}_{total}=-4980J$$
In which of the following case entropy decreases-
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Solid changing to liquid
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Expansion of a gas
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Crystals dissolve
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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 $$300K$$
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$$-69.8$$
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$$-180.75$$
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$$-174.5$$
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$$-139.6$$
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 :
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$$325 kJ/ mol$$
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$$525 kJ/ mol$$
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$$-175 kJ/mol$$
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$$-325 kJ/mol$$
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:
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+ve, -ve
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-ve , +ve
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+ve , +ve
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-ve , -ve
Rank the following substances in order of decreasing heat of combusion (maximum $$\to$$ minimum).
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$$1 > 2 > 4 > 3$$
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$$3 > 4 > 2 > 1$$
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$$2 > 4 > 1 > 3$$
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$$1 > 3 > 2 > 4$$
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.
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7900 bar mL
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8900 bar mL
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9900 bar mL
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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:
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-740.5 kJ/mol
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-741.75 kJ/mol
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-743.0 kJ/mol
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-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
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$$-5kJ/ mol$$
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$$-17.36kJ/ mol$$
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$$-11.2kJ/ mol$$
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$$10.1kJ/mol$$
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 ?
$${\text{FeO}}\left( {\text{s}} \right){\text{ + C}}\left( {{\text{Graphite}}} \right) \to {\text{Fe}}\left( {\text{s}} \right){\text{ + CO}}\left( {\text{g}} \right)$$
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289 K
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668 K
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966 K
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$${{\text{G}}^{\text{o}}}$$ is $${\text{ + ve}}$$. hence the reaction will never be spontaneous
For the process,$${\text{C}}{{\text{O}}_{\text{2}}}\left( s \right) \to \,{\text{C}}{{\text{O}}_{\text{2}}}\left( g \right)$$
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Both $${{\Delta H}}\,\,\,\,$$ and $${{\Delta S}}$$ are negative
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$${{\Delta H}}\,\,\,\,$$ is negative and $${{\Delta S}}$$ is positive
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$${{\Delta H}}\,\,\,\,$$ is +Ve and $${{\Delta S}}$$ is negative
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Both $${{\Delta H}}\,\,\,\,$$ and $${{\Delta S}}$$ are positive
Spontaneity may be observed in following conditions
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$$\triangle H$$ = -ve, $$\triangle S$$ = +ve
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$$\triangle H$$ = +ve, $$\triangle S$$ = +ve
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$$\triangle H$$ = -ve, $$\triangle S$$ = -ve
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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.
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)
$$2C(s) + H_{2}(g) \rightarrow C_{2}H_{2}(g)$$ $$\Delta H = 225 kJ / mol$$
$$2C(s)\rightarrow 2C(g);\Delta H= 1410kJ mol^{-1}$$
$$H_2(g)\rightarrow 2H(g);\Delta H= 330kJ mol^{-1}$$
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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?
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3.4 kcal
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3.4 cal
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3.4 kJ
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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
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298 K
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500 K
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900 K
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none of the above
An endothermic reaction is spontaneous if
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$$\Delta H> T\Delta S$$
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$$\Delta H< T\Delta S$$
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$$\Delta H= T\Delta S$$
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$$T\Delta S=0$$
For the reaction,$$A\rightarrow B,\Delta H=+ve,\Delta S=-ve$$ This reaction is
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non-spontaneous at all temperature
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non-spontaneous at low temperature
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non-spontaneous at high temperature
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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:
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-3400 cal
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3400 cal
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2000 cal
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-2800 cal
The reaction $$2A(g)\rightarrow A_{2}(g)$$, will be spontaneous
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At high temperature
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At low temperature
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At all temperature
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Never at any temperature
The correct thermodynamic conditions for the spontaneous reaction at all temperature is:
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$$\Delta H < 0$$ and $$\Delta S = 0$$
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$$\Delta H > 0$$ and $$\Delta S < 0$$
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$$\Delta H < 0$$ and $$\Delta S > 0$$
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$$\Delta H < 0$$ 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:
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325 kJ/mol
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525 kJ/mol
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-175 kJ/mol
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-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?
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5.87 kJ
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-4.99 kJ
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6.89 kJ
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-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
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-26.14kJ
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44.39kJ
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-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:-
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+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?
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$$+8.43\ J$$
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$$+43\ J$$
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$$+458.43\ J$$
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$$+580.43\ J$$
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
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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_______.
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Spontaneous increase
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Spontaneous decrease
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non-Spontaneous, increase
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non-Spontaneous decrease
For reaction $$ A \rightarrow B $$, $$ \Delta H $$ and $$ \Delta S $$ are positive. the most favourable condition of spontaneous process.
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low temperature
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high temperature
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high concentration
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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
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5.0 V
0%
4.5 V
0%
3.0 V
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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.
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$$70 \%$$ and $$30 \%$$
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$$80 \%$$ and $$20 \%$$
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$$20 \%$$ and $$ 80 \%$$
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$$60 \%$$ 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:
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$$\Delta {H_1} = - \Delta {H_1} + \dfrac{{\Delta {H_2}}}{2} - \dfrac{3}{2}\Delta {H_3}$$
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$$\Delta {H_1} = \Delta {H_1} + \dfrac{{\Delta {H_2}}}{2} - \dfrac{3}{2}\Delta {H_3}$$
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$$\Delta {H_1} = \Delta {H_1} - \dfrac{{\Delta {H_2}}}{2} + \dfrac{3}{2}\Delta {H_3}$$
0%
none of these
An ideal gas expands according to the law $$P^{2}V$$= constant. The internal energy of the gas
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Increases continuously
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Decreases continuously
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Remain constant
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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)$$Na_2S$$
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Only (iv)
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Only (iii) and (iv)
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Only (ii), (iii), (iv)
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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:
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$$ \triangle G = -Ve $$
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$$ \triangle H \simeq 0 $$
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$$ \triangle S = -Ve $$
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$$ \triangle S = +ve $$
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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