MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 16

For a system in equilibrium,

$\Delta G=0$ under conditions of constant:

0%
temperature and pressure
0%
temperature and volume
0%
pressure and volume
0%
energy and volume

Explanation

For a system in equilibrium

$\Delta G=0$ at constant pressure and temperature.

$G(p,T)=U+PV-TS$

$G(p,T)=H-TS$

So, Gibbs's free energy is defined for constant pressure and temperature.

Hence, the correct option is A

At what value of

$\alpha$ will the heat capacity be negative?

0%
$\alpha <-1$
0%
$\alpha <0$
0%
$\alpha >\cfrac { \gamma }{ \gamma -1 }$
0%
$\alpha <\gamma \quad$

What is the heat capacity of the ideal monoatomic gas undergoing the process shown in the figure?

0%
$1.5R$
0%
$2.5R$
0%
$3.5R$
0%
$1.33R$

For which of the following gases, the difference in specific heats at constant pressure and at constant volume is equal to

$0.04545cal$ ?

0%
${N}_{2}$
0%
${N}_{2}O$
0%
$CO$
0%
${CO}_{2}$

Which of the following statement(s) is/are incorrect?

0%
The specific heat capacity of a substance is greater in the solid-state than in liquid state.
0%
The specific heat capacity of a substance is greater in gaseous state than in the liquid state
0%
The latent heat of vaporization of a substance is greater than that of fusion
0%
The internal energy of an ideal gas is a function of its temperature

The standard heat of combustion of propane is

$-2220.1\ kJ/mol$ . The standard heat of vaporization of liquid water is

$44\ kJ/mol$ . What is the

$\triangle H^{o}$ of the reaction :

$C_{3} H_{8} (g)+ 5O_{2} (g) \rightarrow 3CO_{2} (g) + 4H_{2} O (g)$ ?

0%
$-2220.1\ kJ$
0%
$-2044.1\ kJ$
0%
$-2396.1\ kJ$
0%
$-2176.1\ kJ$

For a specific work, on an average a person requires

$5616\ kJ$ of energy. How many kilogram of glucose must be consumed if all the require energy has to be derive from glucose only?

$\Delta H$ for combustion of glucose is

$-2808\ kJ\ mol^{-1}$

0%
$0.720\ kg$
0%
$0.36\ kg$
0%
$0.18\ kg$
0%
$1.0\ kg$

The value of

$\triangle H_{sol}$ of anhydrous copper

$(II)$ sulfate is

$-66.11\ kJ.$ Dissolution of

$1$ mole of blue vitriol (Copper

$(II)$ sulphate pentahydrate) is followed by absorption of

$11.5\ kJ$ of heat. The enthalpy of dehydration of blue vitriol is:

0%
$-77.61\ kJ$
0%
$+77.61\ kJ$
0%
$-54.61\ kJ$
0%
$+54.61\ kJ$

The data below refers to gas phase reaction at constant pressure at

$25^{o}C.$

$CH_{3}- CH_{3} \Rightarrow CH_{3}- CH_{2}+ H; \triangle H_{1} = +420\ kJ\ mol^{-1}$

$CH_{3}- CH_{2} \rightarrow CH_{2} = CH_{2}+ H; \triangle H_{2}+ 168\ kJ\ mol^{-1}$
From these data, the enthalpy change

$\triangle H$ for the reaction :

$2CH_{3} - CH_{2} \rightarrow CH_{3}- CH_{3}+ CH_{2} = CH_{2}$ is

0%
$+250\ kJ$
0%
$+588\ kJ$
0%
$-252\ kJ$
0%
$-588\ kJ$

Geological condition are sometimes so extreme that quantities neglected in normal laboratory experiments take on an overriding importance. For example, consider the formation of diamond under geophysically typical conditions. The density of graphite is

$2.4\ g/cm^{3}$ and that of diamond is

$3.6\ g/cm^{3}$ at a certain temperature and

$500\ kbar$ . By how much does

$\Delta U_{trans}$ differes from

$\Delta H_{trans}$ for the graphite to diamond transition?

0%
$83.33\ kJ/mol$
0%
$0.83\ kJ/mol$
0%
$8.33\times 10^{7}\ kJ/mol$
0%
$83.33\ J/mol$

A quantity that cannot be directly measured is:

0%
heat of formation of $H_{2}O(l)$
0%
heat of formation of $CH_{4}(g)$
0%
latent heat of fusion of ice
0%
heat of combustion of ethyl alcohol

Enthalpy of neutralization of oxalic acid is

$-25.4\ kcal/mol$ using strong base,

$NaOH$ . Enthalpy change for the process:

$H_{2}C_{2}O_{4}(aq)\rightarrow 2H^{+}(aq)+C_{2}O_{4}^{2-}(aq)$ is about

0%
$2.0\ kcal$
0%
$-11.7\ kcal$
0%
$1.0\ kcal$
0%
$4.0\ kcal$

For a spontaneous process:

0%
enthalpy change of the system must be negative
0%
entropy change of the system must be positive
0%
entropy change of the surrounding must be positive
0%
entropy change of the system plus surrounding must be positive

The reaction is ?

0%
spontaneous and exergonic
0%
non-spontaneous and exergonic
0%
spontaneous and endergonic
0%
non-spontaneous and endergonic

When we take acetone in a test tube it feels cold. Which reaction occurs in the process?

0%
endothermic reaction
0%
exothermic reaction
0%
reversible process
0%
adiabatic process

In an ice calorimeter, a chemical reaction is allowed to occur in thermal contact with an ice-water mixture at

$0^{o}C$ . Any heat liberated by the reaction is used to melt some ice; the volume change of the ice-water mixture indicates the amount of melting. When solutions containing

$1.0$ millimole each of

$AgNO_{3}$ and

$NaCl$ were mixed in such a calorimeter, both solution having been pre-cooled to

$0^{o}C$ ,

$0.20\ g$ of ice-melted. Assuming complete reaction in this experiment, what is

$\Delta H$ for the reactions:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ ? Latent heat of fusion of ice at

$0^{o}C$ is

$80\ cal/g$ .

0%
$-16\ kcal$
0%
$+16\ kcal$
0%
$-16\ cal$
0%
$+16\ cal$

Which of the following salts shall cause more cooling when one mole of the salt is dissolved in the same amount of water ? (Integral heat of solution at

$298\ K$ is given for each solute.)

0%
$KNO_{3} ; \triangle H = 35.4\ kJ/mol$
0%
$NaCI; \triangle H = 5.35\ kJ/mol$
0%
$KOH ; \triangle H= -55.6\ kJ/mol$
0%
$HBr ; \triangle H=-83.3\ kJ/mol$

For an ideal solution containing two liquid components A and B, the Gibbs free energy of mixing is minimum, when the molar ratio of the liquids is

0%
1 : 1
0%
1 : 2
0%
1 :10
0%
1 : 1000

Ethyl chlorine is prepared by reaction of ethylene with hydrogen chloride as:

$C_{2}H_{4}(g)+HCl(g)\rightarrow C_{2}H_{5}Cl(g); \Delta H=-72.3\ kJ$

What is the value of

$\Delta U$ (in

$kJ$ ) if

$70\ g$ of ethylene and

$73\ g$ of

$HCl$ are allowed to react?

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

The factor of

$\triangle G$ values is important in metallurgy. The

$\triangle G$ values for the following reactions at

$800^{o}C$ are given as :

$S_{2} (s)+ 2O_{2} (g) \rightarrow 2SO_{2} (g) ; \triangle G=-544\ kJ$

$2Zn(s)+ S_{2} (s) \rightarrow 2Zn S(s); \triangle G= -293\ k J$

$2Zn(s) + O_{2} (g) \rightarrow 2ZnO(s) ; \triangle G= -480\ kJ$
The

$\triangle G$ for the reaction :

$2ZnS(s)+ 3O_{2} (g) \rightarrow 2 ZnO (s) + 2SO_{2}$ will be

0%
$-357\ kJ$
0%
$-731\ kJ$
0%
$-773\ kJ$
0%
$-229\ kJ$

For the reaction,

$A_{\left ( s \right )}+3B_{\left ( g \right )}\rightarrow 4C_{\left ( s \right )}+D_{\left ( l \right )}$ .

$\Delta H$ and

$\Delta U$ are related as:

0%
$\Delta H=\Delta U$
0%
$\Delta H=\Delta U+3RT$
0%
$\Delta H=\Delta U+RT$
0%
$\Delta H=\Delta U-3RT$

Explanation

(D)

$A_{\left ( s \right )}+3B_{\left ( g \right )}\rightarrow 4C_{\left ( s \right )}+D_{\left ( l \right )}$

$\Delta n_{g}=0-3=-3;$

$\Delta H=\Delta E+\Delta n_{g}RT$ {

$\because \Delta E=\Delta U$ }

therefore,

$\Delta H=\Delta U-3RT$

In the reaction for the transition of carbon in the diamond form to carbon in the graphite form,

$\Delta H$ is -453.5 cal. This points out that

0%
graphite is chemically different from diamond
0%
graphite is as stable as diamond
0%
graphite is more stable than diamond
0%
diamond is more stable than graphite

$10 l t$ box contains

$O_3$ and

$O_2$ at equilibrium at

$2000 K$ . The

$\triangle G^o = -534.52 kJ$ at

$8$ atm equilibrium pressure. The following equilibrium is present in the container

$2O_3(g) \rightleftharpoons 3O_2(g)$ . The partial pressure of

$O_3$ will be (ln

$10 = 2.3, R = 8.3 J \;\ mole^{-1} K^{-1})$ .

0%
$8 \times 10^{-6}$
0%
$22.62 \times 10^{-7}$
0%
$9.71 \times 10^{-6}$
0%
$9.71 \times 10^{-2}$

Explanation

$\triangle G^o = -RT$ In

$K=-2.3 \times 2000 \times 8.3 \log K$

$\dfrac{534.52 \times 10^3}{2.3 \times 8.3 \times 2000} = \log K$

$K= 10^{14}$

$2O_3(g) \rightleftharpoons 3O_2(g)$

$K=10^{14}$

$Po_3 <<< Po_2$

$Po_2 + Po_3 = 8$

$Po_2 = 8$ atm

$K = 10^{14} = \dfrac{(Po_2)^3}{(Po_3)^2} = \dfrac{(8)^3}{(Po_3)^2}$

$Po_3 = 22.62 \times 10^{-7}$ atm.

Conditions of standard state used in thermochemistry is?

0%
$0^{\circ}C$ and 1 atm
0%
$20^{\circ}C$ and 1 atm
0%
$25^{\circ}C$ and 1 atm
0%
0 K and 1 atm

$C+O_{2}\rightarrow CO_{2}+94.2 kcal$

$H_{2}+1/2O_{2}\rightarrow H_{2}O+68.3 kcal$

$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O\left ( g \right )+210.8kcal$

Then the possible heat of formation of methane will be

0%
47.3 kcal
0%
20.0 kcal
0%
45.9 kcal
0%
-47.3 kcal

Explanation

Given:

$C+O_{2}\rightarrow CO_{2}; \ \Delta H_f (CO_2) = +94.2 kcal$ (i)

$H_{2}+1/2O_{2}\rightarrow H_{2}O; \ \Delta H_f(H_2O) =+68.3 kcal$ (ii)

$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O\left ( g \right ); \ \Delta H_R=+210.8kcal$ (iii)

$\Delta H_R$ for reaction (iii) can be written as:

$\Delta H = \Delta H_f (CO_2) +2\times \Delta H_f (H_2O) - \Delta H_F (CH_4) -2\times \Delta H_f (O_2)$

$\implies 210.8 = 94.2 + 2 \times 68.3 -\Delta H_f (CH_4) - 2\times 0$

$\Delta H_f (CH_4) = 20 \ kcal$

Hence, Option "B" is the correct answer.

In a thermodynamic system working subtance is ideal gas, its internal energy is in the from of

0%
Kinetic energy only
0%
Kinetic and potential energy
0%
Potential energy
0%
None of these

Which of the following is slow process

0%
Isothermal
0%
Adiabatic
0%
Isobaric
0%
None of these

Gibb's free energy (G) is defined as:

0%
$\Delta G=\Delta H-T\Delta S$
0%
$\Delta G=\Delta H+\frac{T}{\Delta S}$
0%
$\Delta H=\Delta G-T\Delta S$
0%
$\Delta G=\Delta H+T\times C_{P}$

Explanation

Gibbs energy is showing

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

In the Born-Haber cycle for the formation of solid common salt (NaCl), the largest contribution comes from :

0%
the low ionization potential of Na
0%
the high electron affinity of Cl
0%
the low $\triangle H_{vap}$ of Na(S)
0%
the lattice energy

If enthalpy of an overall reaction

$X \rightarrow Y$ along one rout is

$\Delta H$ and

$\Delta H_{1}, \Delta H_{2}, \Delta H_{3},....$ representing enthalpies of reactions leading to same product Y then

$\Delta H$ is

0%
$\Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3}...$
0%
$\Delta H = \Delta H_{1} \times \Delta H_{2} \times \Delta H_{3}...$
0%
$\Delta H = \Delta H_{1} + \Delta H_{2} - \Delta H_{3}....$
0%
$\Delta H = \dfrac{\Delta H_{1} \times \Delta H_{2} \times \Delta H_{3}}{2}....$

Explanation

According to Hess's law,

$\Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3}......$

Where

$\Delta H_{1}, \Delta H_{2} , \Delta H_{3} ,$ etc. are enthalpies of various steps leading to final same product.

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.

Explanation

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

$\text{Hess's law of constant heat summation.}$

A reaction proceeds through two paths I and II to convert

$X \rightarrow Z$
What is the correct relationship between

$Q, Q_{1}$ and

$Q_{2}$ ?

0%
$Q = Q_{1} \times Q_{2}$
0%
$Q = Q_{1} + Q_{2}$
0%
$Q = Q_{1} - Q_{2}$
0%
$Q = Q_{1} / Q_{2}$

Explanation

Hess's Law of Constant Heat Summation (or just 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.

So,

$Q = Q_{1}+Q_{2}$

A gaseous system is initially characterised by 500 mL volume and 1 atm pressure at 298 K. This system is allowed to do work as

(i) In isobaric conditions it expands to 800 mL resulting a decrease in pressure and temperature to 0.6 atm and 273 K respectively.

(ii) In adiabatic conditions it is allowed to expand upto 800 mL and results a decrease in pressure and temperature to 0.6 atm and 273 K respectively.

If Glibbs energy change in (i)

$\Delta G_{a}$ and in is

$\Delta G_{b}$ then what will be the ratio of

$\dfrac{\Delta G_{a}}{\Delta G_{b}}?$

0%
$0$
0%
$1$
0%
between $0-1$
0%
$> 1$

Explanation

$\dfrac{\Delta G_{a}}{\Delta G_{b}}=1$ as Gibbs energy is a state function and initial and final states are same in (i) and (ii)

Option B is correct.

Which thermochemical process is shown by the following figure?

0%
Standard enthalpy of a reaction
0%
Born - Haber cycle of lattice enthalpy
0%
Hess's law of constant heat summation
0%
Standard enthalpy of a solution .

Explanation

The thermochemical process shown by the figure is: Hess's law of constant heat summation.

$X \overset{\Delta H}{\rightarrow} Y$

$X\overset{\Delta H_{1}}{\rightarrow} P\overset{\Delta H_{2}}{\rightarrow} Q\overset{\Delta H_{3}}{\rightarrow} Y$

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

Read the following statements regarding the spontaneity of a process and mark the appropriate choice.
(i) When enthalpy factor is absent then randomness factor decides spontaneity of a process.
(ii) When randomness factor is absent then enthalpy factor decides spontaneity of a process.
(iii) When both the factors take place simultaneously the magnitude of both factors decides spontaneity of a process.

0%
Statements (i) and (ii) are correct and (iii) is incorrect.
0%
Statements (iii) and (i) are correct and (ii) is incorrect.
0%
Statements (i), (ii) and (iii) are correct.
0%
Statements (i), (ii) and (iii) are incorrect.

Explanation

We know,

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

Now,

1.

$\Delta G=\Delta H$ when

$\Delta S=0.$

$\Delta G=T\Delta S$ when

$\Delta H=0.$

3.

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

$\Delta H\neq0\neq \Delta S$

At dynamic equilibrium, the reactions on both sides occur at the same rate and the mass on both sides of the equilibrium does not undergo any change. This condition can be achieved only when the value of

$\Delta G$ is:

0%
$-1$
0%
$+1$
0%
$+2$
0%
$0$

Which of the following conditions may lead to a non-spontaneous change?

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

Using the listed [

$\Delta {G}_{f}^{o}$ values] calculate

$\Delta {G}^{o}$ for the reaction:

$3{H}_{2}S(g)[-33.6]+2H{NO}_{3}(l)[-80.6]\rightarrow 2NO(g)[+86.6]+4{H}_{2}O(l)[-237.1]+3S(s)[0.0]$

0%
$-513.0$
0%
$-1037.0$
0%
$+433.4$
0%
$+225.0$

Which of the following conditions will always lead to non-spontaneous change?

0%
$\Delta H$ and $\Delta S$ both +ve
0%
$\Delta H$ is -ve and $\Delta S$ is +ve
0%
$\Delta H$ and $\Delta S$ both -ve
0%
$\Delta H$ is +ve and $\Delta S$ is -ve

Which of the following options is/are correct?

0%
$\left[ \cfrac { \partial \ln { { K }_{ p } } }{ \partial T } \right] =\cfrac { \Delta { H }^{ o } }{ R{ T }^{ 2 } }$
0%
$\cfrac { \partial \ln { { K }_{ } } }{ \partial T } =\cfrac{{E}_{a}}{R{T}^{2}}$
0%
$\left[ \cfrac { \partial \ln { { K }_{ p} } }{ \partial T } \right] =\cfrac { \Delta { U }^{ } }{ R{ T }^{ 2 } }$
0%
All of these

The free energy change

$\Delta G=0$ , when

0%
the system is at equilibrium
0%
catalyst is added
0%
reactants are initially mixed thoroughly
0%
the reactants are completely consumed

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

0%
be reduced to half
0%
be doubled
0%
remain constant
0%
be increased four times

The heat required to raise the temperature of a body by

$1\ K$ is called :

0%
specific heat
0%
thermal capacity
0%
water equivalent
0%
molar heat capacity

Calculate the enthalpy of formation of

$I_2O_5(s)$ from the following data:
i)

$I_2O_5(s) + H_2O(l)\rightarrow2HIO_3(aq)$ ;

$\Delta H = +4.0kJ$
ii)

$KI(aq) + 3HClO(aq) \rightarrow HIO_3(aq) + 2HCl(aq) + KCl(aq)$ ;

$\Delta H = -322.0kJ$
iii)

$NaOH(aq) +HClO(aq) \rightarrow NaOCl(aq) + H_2O(l)$ ;

$\Delta H = -44.0kJ$
iv)

$NaOH (aq) +HCl(aq)\rightarrow NaCl(aq) +H_2O(l)$ ;

$\Delta H = -57.0kJ$
v)

$2NaOH(aq) +Cl_2(g)\rightarrow NaOCl(aq) +NaCl(aq)+ H_2O(l)$ ;

$\Delta H = -100.0kJ$
vi)

$2KI(aq) +Cl_2(g) \rightarrow 2KCl(aq) +I_2(s)$ ;

$\Delta H = -224.0kJ$
vii)

$H_2(g) +1/2O_2(g) \rightarrow H_2O(l)$ ;

$\Delta H = -285.0kJ$
viii)

$1/2H_2(g) +1/2Cl_2(g)\rightarrow HCl(g)$ ;

$\Delta H = -92.0kJ$
ix)

$HCl(g) +aq \rightarrow HCl(aq)$ ;

$\Delta H = -75.0kJ$

0%
$-173.0\, kJ$
0%
$+173.0\, kJ$
0%
$+154.0 \,kJ$
0%
$-154.0 \,kJ$

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 16 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 16 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.