Assuming that water vapour is an ideal gas, the internal energy change

$(\Delta U)$ when

$1 mol$ of water is vapourised at

$1 bar$ pressure and

$100 ^ { \circ } C .$ (Given: Molar enthalpy of vapourisation of water at

$1 bar$ and

$373 K = 41 kJ mol ^ { -1 }$ and

$R = 8.3 \mathrm { Jmol } ^ { - 1 } \mathrm { K } ^ { - 1 } \text { will be } )$

0%
$4.100 \mathrm { kJ } \mathrm { mol } ^ { - 1 }$
0%
$3.7904 \mathrm { kJ } \mathrm { mol } ^ { - 1 }$
0%
$37.904 \mathrm { kJ } \mathrm { mol } ^ { - 1 }$
0%
$41.00 \mathrm { kJ } \mathrm { mol } ^ { - 1 }$

Which one of the following is correct?

0%
$-\Delta G = \Delta H - T \Delta S$
0%
$-\Delta H = \Delta G - T \Delta S$
0%
$-\Delta S = \dfrac{1}{T}[\Delta G - \Delta H]$
0%
$-\Delta S = \dfrac{1}{T}[\Delta H - \Delta G]$

Explanation

The relation between Gibb's free energy, enthalpy, and entropy is given as:

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

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

$\Delta S=\dfrac1T[\Delta H-\Delta G]$

$-\Delta S=\dfrac1T[\Delta G-\Delta H]$

The internal energy of a gas in an adiabatic process is given by

$U=a+bPV$ find y:-

0%
$\dfrac { a+1 }{ a }$
0%
$\dfrac { b+1 }{ b }$
0%
$\dfrac { b+1 }{ a }$
0%
$\dfrac { a }{ b+1 }$

The

$\Delta H$ and

$\Delta S$ for a reaction at one atmospheric pressure are

$+30.558\ KJ$ and

$+0.066\ KJ$ respectively. The temperature at which the free energy changes will be zero and below this temperature, the nature of the reaction would be

0%
$483\ K,spontaneous$
0%
$443\ K,non-spontaneous$
0%
$463\ K,spontaneous$
0%
$463\ K,non-spontaneous$

Consider the following reaction .

${C_6}{H_6}\left( \ell \right) + {{15} \over 2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( g \right)$
Signs of

$\Delta H,$

$\Delta S$

$\Delta G$ for the above reaction will be;

0%
$+ ,\,\, - ,\,\, +$
0%
$- ,\,\, + ,\,\, +$
0%
$- ,\,\, + ,\,\, -$
0%
$+ ,\,\, + ,\,\, -$

The standard reduction potentials for

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

$Fe^{2+}$ are -0.76, -0.23 and -0.44V respectively. The reduction

$X +Y^{2+} \rightarrow X^{2+} +Y$ will be spontaneous when :

0%
X = Ni, Y = Zn
0%
X=Fe, Y = Zn
0%
X = Zn, Y = Ni
0%
X = Ni, Y= Fe

The internal energy of gases

$He, O_{2}$ and

$NH_{3}$ are plotted against the absolute temperature. The respective graphs

$1, 2$ and

$3$ are of?

0%
$He, O_{2}$ and $NH_{3}$
0%
$NH_{3}, H_{2}$ and $O_{2}$
0%
$NH_{3}, O_{2}$ and $He$
0%
$O_{2}, He$ and $NH_{3}$

For the reaction at 298K

$CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g), \Delta H^{o}=178.3kJ,\Delta S^{o}=160Jk^{-1}$

Select correct statement

$(s)$ :

0%
The reaction is sponteneously at this temperature $(298 K)$
0%
If temperature is decrease forward reaction is fevoured
0%
The reaction is spontaneously in forward direction onlt at temperature above $1000K$
0%
The reaction is spontaneously in forward direction onlt at temperature above $1.114K$

Consider the following reaction

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\left( \ell \right){\text{ + }}\frac{{{\text{15}}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{6C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O}}(g)$
Signs of

$\Delta {\text{H,}}\,\,\Delta {\text{S}}\,{\text{and}}\,\Delta {\text{G}}$ for the above reaction will be:

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

$\triangle G$ of the conversion of 2 mol of

$C_6H_6(I) at 80^oC$ (normal boiling point) to vapour at the same temperature and a pressure of 0.2 atm is:

0%
-9.44 Kcl/mol
0%
-2.27 Kcal/mol
0%
-1.135 Kcal/mol
0%
zero

Heat of

$30 kcal$ is supplied to a system and

$4200 J$ of external work is done on the system so that its volume decreases at constant pressure. What is the change in its internal energy.

$=(J=4200J/kcal)$

0%
$1.302\times 10^5J$
0%
$2.302\times 10^5J$
0%
$3.302\times 10^5J$
0%
$4.302\times 10^5J$

Explanation

$\Delta Q = 30 \times 4200$

$= 126000J$

$\Delta W = - 4200J$

$\therefore \Delta u = \Delta Q - \Delta W$

$= 126000 + 4200J$

$= 1.302 \times {10^5}J$

Hence,

option

$(A)$ is correct answer.

A diatomic gas is heated at constant at constant pressure. What fraction of the heat energy is used to increase the internal energy ?

0%
$\dfrac 3 5$
0%
$\dfrac 3 7$
0%
$\dfrac 5 7$
0%
$\dfrac 5 9$

The enthaply change on freezing of

$1\ mol$ of water at

$10^ {o}C$ to ice at

$-10^ {o}C$ is:

0%
$6.56\ kJ\ mol^ {-1}$
0%
$5.81\ kJ\ mol^ {-1}$
0%
$6.00\ kJ\ mol^ {-1}$
0%
$5.64\ kJ\ mol^ {-1}$

Given that:
(i)

$C$ (graphite)

$+O_2(g)\rightarrow CO_2(g)$ ;

$\quad \Delta _rH^o=x\ kJ mol^{-1}$ .
(ii)

$C$ (graphite)

$+\dfrac{1}{2}O_2(g)\rightarrow CO(g)$ ;

$\quad \Delta _rH^o=y\ kJ mol^{-1}$
(iii)

$CO(g)+\dfrac{1}{2}O_2(g)\rightarrow CO_2(g)$ ;

$\quad \Delta _rH^o=z\ kJ mol^{-1}$

Based on the given thermochemical equations, find out which one of the following algebraic relationships is correct?

0%
$z=x+y$
0%
$x=y-z$
0%
$x=y+z$
0%
$y=2z-x$

Explanation

$C_{(graphite)}+O_2(g)\rightarrow CO_2(g)\Delta_rH^o=xkJ/mol$ .

$(1)$

$C_{(graphite)}+\dfrac{1}{2}O_2(g)\rightarrow CO(g)\Delta_rH^o=ykJ/mol$ ..

$(2)$

$CO(g)+\dfrac{1}{2}O_2(g)\rightarrow CO_2(g)\Delta_rH^o=zkJ/mol$ .

$(3)$
Now,

$(1)=(2)+(3)$

$x=y+z$ .

The sole criterion for the spontaneity of a process is

0%
tendency to acquire minimum energy
0%
Tendency to acquire maximum randomness
0%
Tendency to acquire minimum energy and maximum randomness
0%
tendency to acquire maximum stability

The internal energy of an ideal gas increases when it

0%
Expands
0%
Is compressed
0%
Is first expanded and then compressed
0%
None of these

A reaction is spontaneous at low temperature but non-spontaneous at high temperature. Which of the following is true for the reaction?

0%
$\triangle H > 0,\triangle S > 0$
0%
$\triangle H < 0,\triangle S > 0$
0%
$\triangle H > 0,\triangle S=0$
0%
$\triangle H < 0,\triangle S < 0$
0%
$\triangle H = 0,\triangle S < 0$

$27^oC$ the reaction,

$C_6H_{6(1)} + \dfrac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} +3H_2O_{(I)}$
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%
$\Delta H\ >\ 0\ and\ T\Delta S\ <\ 0$

In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas molecules gives out 20 J of heat and 10 J of work is done on the gas. If the initial internal energy of the gas was 40 J, then the final internal energy will be

0%
$30 J$
0%
$20 J$
0%
$60 J$
0%
$40 J$

A process has

$\Delta H = 200 \,J \,mol^{-1}$ and

$\Delta S = 40 \,JK \,mol^{-1}$ . Out of the values given above which the process will be sponteneous:

0%
$5 \,K$
0%
$4 \,K$
0%
$20 \,K$
0%
$12 \,K$

Explanation

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

$T = \dfrac{\Delta H}{\Delta S} = \dfrac{200}{40} = 5K$

A thermodynamical process is shown in the figure. The pressure and volumes corresponding to some points in the figure are

${ P }_{ A } = 3 \times {10}^{4} Pa$

${ V }_{A} = 2 \times {10}^{-3} { m }^{3}$

${ P }_{ B } = 8 \times {10}^{4} Pa$

${ V }_{D} = 5 \times {10}^{-3} { m }^{3}$ In the process

$AB$ ,

$600 J$ of heat is added to the system and in process

$BC 200 J$ of heat is added to the system. The change in internal energy of the system in process

$AC$ would be

0%
$560 J$
0%
$800 J$
0%
$600 J$
0%
$640 J$

Ethyl chloride

$(C_{2}H_{5}Cl)$ is prepared by:-

$C_{2}H_{4(g)}+HCl_{(g)}\rightarrow C_{2}H_{5}Cl_{(g)};\ \ \Delta H=-72.3 kJ$ if 98 g of

$C_{2}H_{4}$ and 109.5 g

$HCl$ are used at 300 K, then find

$\Delta E$ .

0%
-64.8 kJ
0%
-190 kJ
0%
209.4 kJ
0%
-209.4 kJ

A gas is compressed at a constant pressure of 50

$N/m^{ 2 }$ from a volume of

$10m^{ 3 }$ to a volume of

$4m^{ 3 }$ . Energy of 100 J is then added to the gas by heating. Its internal energy is

0%
increased by 400 J
0%
increased by 200 J
0%
increased by 100 J
0%
decreased by 400 J

Which relationship is incorrect? (for reversible isothermal process)

0%
$\Delta H=\Delta U+\Delta n_{g}$ RT(for reaction)
0%
$\Delta G= T \Delta S$
0%
$\Delta G^{\circ}=-2.303$ RT log K
0%
W= +2.303 nRT log $\frac{V_{1}}{V_{2}}$

The amount of heat energy required to raise the temperature

${\text{1}}\,{\text{g}}$ of Helium at

${\text{NTP}}$ , from

${\text{T,K}}$ to

${{\text{T}}_2}{\text{K}}$ is

0%
$\frac{3}{2}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$
0%
$\frac{3}{4}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$
0%
$\frac{{\text{3}}}{{\text{4}}}{{\text{N}}_{\text{a}}}{{\text{k}}_{\text{B}}}\left( {\frac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}} \right)$
0%
$\frac{{\text{3}}}{8}{{\text{N}}_{\text{a}}}{{\text{k}}_{\text{B}}}\left( {{T_2} - {T_1}} \right)$

The amount of heat required to raise the temperature of 1 mole diatomic gas by

$1^0C$ a constant pressure is 60 cal. The amount of heat which goes as internal energy of the gas is nearly

0%
60 cal
0%
30 cal
0%
42.6 cal
0%
49.8 cal

Energy required to dissociate, 16 g oxygen gas into free atom is x kJ. The heat of atomisation of oxygen is :-

0%
x/2
0%
2x
0%
x
0%
16

A process has

$\Delta H=200 J mol^{-1}$ and

$\Delta S=40 JK^{-1} mol^{-1}$ . Out of the values given below, choose the minimum temperature above which the process will be spontaneous?

0%
4 K
0%
20 K
0%
5 K
0%
12 k

For a spontaneous process, if entropy and volume are constant, the internal energy system must

0%
Increase
0%
Decrease
0%
Remain constant
0%
Be zero

A process is said to be spontaneous if

0%
$\triangle G = -ve$
0%
$\triangle G^{\circ} = -ve$
0%
$\triangle H = -ve$
0%
$\triangle H^{\circ} = -ve$

A reaction has

$\triangle H = -33\ kJ$ and

$\triangle S = +58\ J/K$ . This reaction would be:

0%
spontaneous below a certain temperature
0%
non-spontaneous at all temperature
0%
spontaneous above a certain temperature
0%
spontaneous at all temperature

A body cools from

$71^0C$ to

$69^0C$ in

$4$ minutes. If the surrounding temperature is constant at

$20^0C$ . Time taken by the body to cool from

$61^0C$ to

$59^0C$ is

0%
$5$ minutes
0%
$4$ minutes
0%
$6$ minutes
0%
$3$ minutes

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is ______________________.

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

An ideal monoatomic gas is taken through a process in which

$d Q = 2 d U$ . The molar heat capacity of the gas for the process is

0%
$R$
0%
$2R$
0%
$3R$
0%
$4R$

Which of the following integrals depends only on the initial and final states of a thermodynamic system (i.e., independent of the path of transformation)?

0%
$\int P d V$
0%
$\int d Q$
0%
$\int T^2 d S$
0%
none of these

The P-V diagrams of two difference masses

$m_1$ and

$m_2$ for an ideal gas at constant temperature

$T$ is given in the figure below.

0%
$m_1 = m_2$
0%
$m_1 < m_2$
0%
$m_1 > m_2$
0%
Data insufficient

Correct desending order of deprotonation in the following compound:

0%
e > d > c > b > a
0%
e > d > c > a > b
0%
e > c > b > d > a
0%
e > c > b > a > d

Using the Gibbs energy change ,

${ \Delta G }^{ \circ }=+63.3 kJ,$ for the following reaction,

${ Ag }_{ 2 }{ CO }_{ 3 }\rightleftharpoons 2Ag\left( sq \right) +{ CO }^{ 2- }_{ 3 }\left( aq \right)$ the

${ K }_{ sp }of{ Ag }_{ 2 }{ CO }_{ 3 }\left( s \right)$ in water at

${ 25 }^{ \circ }C$ is:-

$\left( R={ 8.314\quad j=J\quad k }^{ -1 }{ mol }^{ -1 } \right)$

0%
${ 3.2\times 10 }^{ -26 }$
0%
${ 8.0\times 10 }^{ -12 }$
0%
${ 2.9\times 10 }^{ -3 }$
0%
${ 7.9\times 10 }^{ -2 }$

A gas mixture consists of

$2$ moles of O, and

$4$ moles of Ar at a temperature T. Neglecting all vibrational moles, the total internal energy of the system is

0%
$4RT$
0%
$15RT$
0%
$9RT$
0%
$11RT$

Hess's law states that:

0%
The standard enthalpy of an overall reaction is the sum of the enthalpy changes in individual reactions.
0%
Enthalpy of formation of a compound is same as the enthalpy of decomposition of the compound into constituent elements, but with opposite sign.
0%
At constant temperature, the pressure of a gas is inversely proportional to its volume
0%
The mass of a gas dissolve per liter of a solvent is proportional to the pressure of the gas in equilibrium with the solution.

Which of the following are spontaneous?

0%
dissolution of sugar
0%
separation of Ar and Kr from their mixture
0%
spreading of fragrance when a bottle of perfume is opened
0%
Flow of heat from cold object to hot object
0%
heat transfer from ice to room at $25^0 C$

One mole of an ideal monatomic gas undergoes a process described by the equation

$PV^{3}$ = constant. The heat capacity of the gas during this process is:

0%
$R$
0%
$\dfrac{3}{2R}$
0%
$\dfrac{5}{2R}$
0%
${2R}$

Assertion : A reaction which is spontaneous and accompanied by decrease of randomness must be exothermic.
Reason : All exothermic reactions are accompanied by decrease of randomness.

0%
both assertion and reason are true and the reason is the correct explanation of assertion.
0%
both assertion and reason are true but the reason is not the correct explanation of assertion.
0%
assertion is true but reason is false.
0%
both assertion and reason are false

A reaction A(g) +B(g)

$\rightarrow$ C(g) +D(g).

$\Delta H$ = (-) ve is found to have positive entropy change. the reaction will be:

0%
Spontaneous at high temperature
0%
Spontaneous only at low temperature
0%
Not spontaneous at any temperature
0%
Spontaneous at any temperature

In which case, process will be spontaneous at all temperatures?

0%
$\Delta H<0,$ $\Delta S>0$
0%
$\Delta H>0,$ $\Delta S>0$
0%
$\Delta H<0,$ $\Delta S<0$
0%
$\Delta H>0,$ $\Delta S<0$

Explanation

Solution:- (A)

$\Delta{H} < 0$ and

$\Delta{S} > 0$

For spontaneous process

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

$\because \Delta H<0$

$\Delta S>0$

Consider the following reaction occurring in an automobile:

$2C_8H_{18(g)} + 25 O_{2(g)} \rightarrow 16 CO_{2(g)} + 18H_2O_{(g)}$ , the sign of

$\Delta H, \ \Delta S$ and

$\Delta G$ would be:

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

$100g$ of water is heated from

${30}^{o}C$ to

${50}^{o}C$ . Ignoring the slight expansion of water, the change in its internal energy is (Specific heat of water is

$4200J$

${kg}^{-1}$

${K}^{-1}$ )

0%
$4.2kJ$
0%
$84kJ$
0%
$2.1kJ$
0%
$8.4kJ$

The heat energy of

$743J$ is needed to raise the temperature of

$5$ moles of an ideal gas by

$2K$ at constant pressure. How much heat energy is needed to raise the temperature of the same mass of the gas by

$2K$ at constant volume?

0%
$826J$
0%
$743J$
0%
$660J$
0%
$600J$

Explanation

as we know,

$(Q)_P=nC_PΔT=\Delta H$

$C_P−C_V=R⇒C_V=C_P−R,\ (Q)_V=nC_VΔT=\Delta U$

At constant pressure,

$\Delta H=nC_PΔT=743J$

At constant volume,

$\Delta U=nC_vΔT=n(C_p−R)\Delta T=nC_PΔT-nRΔT$

$\Delta U=743−(5×8.3×2)$

$\Delta U=660J$

Option C is correct.

A student mixed 25.0cm3 of 4.00 moldm–3 hydrochloric acid with an equal volume of 4.00moldm–3 sodium hydroxide.?

The initial temperature of both solutions was 15.0°C. The
maximum temperature recorded was 30.0°C.
Using these results, what is the enthalpy change of neutralisation of hydrochloric acid?

0%
$-62.7kJ$ ${mol}^{-1}$
0%
$-31.4kJ$ ${mol}^{-1}$
0%
$-15.7kJ$ ${mol}^{-1}$
0%
$-3.14kJ$ ${mol}^{-1}$

Any process will be spontaneous at constant pressure and temperature when:

0%
$\Delta { S }_{ system }=+ve$
0%
$\Delta { S }_{ univ. }=+ve$
0%
$\Delta { G }_{ sys }=-ve$
0%
$\Delta { G }_{ univ. }=+ve$

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 15 - MCQExams.com

A student mixed 25.0cm3 of 4.00 moldm–3 hydrochloric acid with an equal volume of 4.00moldm–3 sodium hydroxide.?

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 15 - MCQExams.com

A student mixed 25.0cm3 of 4.00 moldm–3 hydrochloric acid with an equal volume of 4.00moldm–3 sodium hydroxide.?

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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