MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 2

Energy hidden in a definite quantity of substance is:

0%
Enthalpy
0%
Internal energy
0%
Free energy
0%
Entropy

Hess's law of constant heat summation is based on

0%
E = mc $^{2}$
0%
conservation of mass
0%
first law of thermodynamics
0%
E = h $\nu$

Explanation

Hint: Hess's law of constant heat summation states that " Regardless of multiple stages, total enthalpy changes is the sum of all the changes."

Correct answer: Option

$C$

$\bf{Explanation \ for \ Correct \ Option}:$

Mathematically Hess law can be represented as;
$\Delta H_{net} = \sum \Delta H_r$
Where, $\Delta H_{net}$ = Net enthalpy change
$\Delta H_r$ =Sum of enthalpy change of reactions
So, Enthalpy change is a state function as it depends on the final and initial stage and not the path followed by the system.
So, It resembles internal energy as internal energy is also a state function.
The first law of thermodynamics states that " Some part of internal energy of the system is used to heat the system while another part is used for work done."

So, Hess's law of constant heat summation is based on the first law of thermodynamics as both suggest the conversation of energy.

$\bf{Explanation \ for \ Incorrect \ Options}$

The statement $E = mc^2$ suggests the relationship between mass and energy and Hess's law deals with energy only.
In Hess law conservation of energy takes place and not conservation of mass.
$E = h \mu$ suggest that energy is quantized so it does not represent Hess law.

Regarding a thermochemical equation, wrong statement is

0%
It tells about the physical states of reactants and products
0%
It tells whether the reaction is exothermic or endothermic
0%
It tells about the allotropic form (if any) of the reactant
0%
It tells whether the reaction is possible or not

All natural processes are:

0%
Spontaneous
0%
Non- spontaneous
0%
Exothermic
0%
Endothermic

The heat change associated with reactions at constant volume is due to the difference in which property of the reactants and the products ?

0%
internal energy
0%
enthalpy
0%
heat capacity
0%
free energy

Explanation

First Law of Thermodynamics dictates that

$\Delta Q = \Delta U + \Delta W$

At constant V,

$\Delta W = 0$

So,

$\Delta Q = \Delta U$

So at constant volume, heat change takes place due to change in internal energy of the system

Hess’s law deals with:

0%
heat changes in a chemical reaction.
0%
rate of reaction
0%
equilibrium constant
0%
influence of pressure on volume of a gas

Which type of molecular motion does contribute towards internal energy for an ideal mono-atomic gas?

0%
Translational
0%
Rotational
0%
Vibrational
0%
All of the above

Explanation

$U =\dfrac{f}{2}N \ k \ T$ where f = number of degrees of freedom

In an ideal mono-atomic gas, the internal energy is contributed only due to the translational kinetic energy.

For example if we take He or Ne, the atoms are not bound to each other and hence they do not vibrate. Rotational energy is neglected since their atomic moment of inertia is minimal and higher energy excitation electronically also is not possible (except at very high temperatures).

Thus, the internal energy changes in an ideal mono atomic gas are purely translational and there are 3 degrees of freedom. In other words translation can take place in 3D space i.e. x, y and z directions.

If there are N atoms in the gas, the total energy U is given by

$U =\dfrac{3}{2}N \ k \ T$

One mole of a monoatomic gas is heated at a constant pressure of

$1atm$ from

$0K$ to

$100K$ . If the gas constant

$R=8.32J{mol}^{-1}{K}^{-1}$ , the change in the internal energy of the gas is approximately

0%
$2.3J$
0%
$46J$
0%
$8.67\times{10}^{3}J$
0%
$1.25\times{10}^{3}J$

When a solid melts, there is

i) increase in enthalpy

ii) decrease in internal energy

iii) decrease in enthalpy

0%
all are correct
0%
i only correct
0%
ii only correct
0%
iii only correct

Explanation

(solid)

$\rightarrow$ liquid

increase in enthalpy

increase in internal energy

increase in entropy

What is the free energy change, $'\Delta G'$ When 1.0 mole of water at 100 $^{0}$ C and 1atm pressure is converted into steam at 100 $^{0}$ C and 1atm pressure

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

Explanation

water

$\rightleftharpoons$ steam

water is in equilibrium with steam at 1 atm pressure and

$100^{\circ}$ C

for a process at equilibrium

$\Delta G=O$

$\Delta H_{f}=$ -98.2 K.Cal/mole

$S_{Na} =$ 36 K.Cal/mole

$I_{Na} =$ 118.5 K.Cal/mole

$\dfrac{1}{2}D_{Cl_{2}}=$ 29 K.Cal/mole

$U_{NaCl}=$ -184.2 K.Cal/mole
From the data given below for

$NaCl$ , the electron affinity of chlorine

$[-E_{a}]$ is:

0%
-97.5 K.cal/ mole
0%
-108 K.cal/mole
0%
-75 K.cal/mole
0%
-128 K.cal/mole

Explanation

From Born-Haber Cycle,
We have,

$\Delta H_f = S_{Na} + I_{Na} + \frac {1}{2} D_{Cl_2} + E_a + U_{NaCl}$

$-98.2 = 36 + 118.5 + 29 + (-184.2) + E_a$

$E_a = -97.5 K. Cal/mole$

Born - Haber cycle may be used

0%
to find out electron affinity of non-metal atoms
0%
to find out lattice energy of the ionic compounds
0%
for the preparation of ammonia in industries
0%
both A and B

Which of the following relations is not correct?

0%
$G=H-TS$
0%
$\Delta G_{system} = \Delta H _{system} T \Delta S_{system}$
0%
$T \Delta S_{system} = \Delta H_{system} -\Delta G_{system}$
0%
$\Delta H_{system} =\Delta G _{system} + T\Delta S_{system}$

Explanation

$G$ =

$H - TS$

$\Rightarrow \Delta G_{system}=\Delta H_{system}-T\Delta S_{system}$

$\Rightarrow T\Delta S_{system}=\Delta H_{system}-\Delta G_{system}$

$\Rightarrow \Delta H_{system}=\Delta G_{system}+T\Delta S_{system}$

Thus, option

$B$ is not correct relation.

Hence, the correct option is

$B$

Gibbs energy change $\Delta$ G is related to equilibrium constant K as:

0%
$\Delta G^{0}$ = $- RT lnk$
0%
$\Delta G^{0}$ = $RT lnk$
0%
$lnk$ = $-\frac{RT}{\Delta G^{0}}$
0%
$lnk$ = $\frac{\Delta G^{0}}{RT}$

Explanation

$\Delta G^{o}=-RT$ Ink

Gibbs energy change

$\Delta G$ is related to equilibrium constant K as above

Which of the following relation is incorrect?

0%
$\Delta S=\Delta H$ + $T\Delta G$
0%
$\Delta S=\frac{\Delta H-\Delta G}{T }$
0%
$– \Delta G = - W_{non\ exp.}$
0%
$W_{PV} = –P \Delta V$

Explanation

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

$\therefore \Delta S=\frac{\Delta H-\Delta G}{T}$

$\Delta G=-W_{non\ exp.}$

Gibbs free energy is equal to useful work for a spontaneous process

$P-V$ work

$=-P\Delta V$

In a chemical reaction, $\Delta H$ =150KJ and $\Delta S$ = 100JK $^{-1}$ at 300K. $\Delta$ G would be ________.

0%
Zero
0%
300KJ
0%
330KJ
0%
120KJ

Explanation

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

$=150KJ-300\times 100\times 10^{-3}KJ$

$=150-30KJ$

$\Delta G=120KJ$

Correct relation among the following.

0%
$\Delta G_{system} = \Delta S_{total}$
0%
$\Delta G_{system}=\Delta H_{system}-T\Delta S_{system}$
0%
$\Delta$ G = $\Delta$ H + T $\Delta$ S
0%
$\Delta S=\frac{\Delta G+\Delta H}{\Delta T}$

Explanation

$\Delta G_{system}=T\Delta S_{total}$

$\Delta G_{system}=\Delta H_{system}-T\Delta S_{system}$
If an isolated system (Q

$=$ 0) is at constant pressure (Q

$=\Delta H$ ) then

$\Delta H=0$
Therefore, the Gibbs free energy of an isolated system is

$\Delta G_{system}=-T\Delta S_{total}$

Gibbs -Helmholtz equation is______

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

Explanation

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

is the Gibbs Helmholtz equation.

Option A is correct.

Which of the following statements is incorrect?

0%
It is not possible to compress a gas at a temperature below $T_c$
0%
At a temperature below $T_c$ , the molecules are close enough for the attractive forces to act, and condensation occur
0%
No condensation takes place above $T_c$
0%
The kinetic energy of the gas molecules is higher above $T_c$ , and the attraction between them decreases

Explanation

Critical temperature of the substance is the temperature at a substance's critical point.
Above

$T_c$ we cannot liquify gas keep pressure constant although on

$\uparrow$ pressure we can liquify.

Identify the correct statement for change of Gibbs energy for a system (

$\Delta$ G

$_{system}$ ) at constant temperature and pressure:

0%
If $\Delta$ G $_{system}$ > 0, the process is spontaneous
0%
If $\Delta$ G $_{system}$ = 0, the system has attained equilibrium
0%
If $\Delta$ G $_{system}$ = 0, the system is still moving in a particular direction
0%
If $\Delta$ G $_{system}$ < 0, the process is not spontaneous

Explanation

At equilibrium, the entropy of system is maximum. Therefore change in entropy i.e;

$\triangle S=0$

Process	Sign of $\triangle S$	Sign of $\triangle G$
Spontaneous	Positive, $\triangle S_{total} >0$	Negative, $\triangle G <0$
Non-Spontaneous	Negative, $\triangle S_{total} < 0$	Positive, $\triangle G>0$
Equilibrium	$\triangle S_{total}=0$	$\triangle G=0$

The equilibrium partial pressure of

$N_2, H_2$ and

$NH_3$ are

$4, 4$ and

$8$ atm respectively. The value of

$K_p$ for the Haber's process in

$atm^{-1}$ is:

0%
$1/4$
0%
$1/2$
0%
$4$
0%
$2$

Cell reaction is spontaneous when :

0%
G is negative
0%
G is positive
0%
E is positive
0%
E is negative

Explanation

Hint: The Gibbs free energy is related to the spontaneity of a reaction.

Explanation: If the Gibbs free energy is negative then the reaction is spontaneous.

$\Delta {G^ \circ } = - nF{E^ \circ }$

For $\Delta {G^ \circ } < 0$ , the reaction, ${E^ \circ }_{cell}$ should be positive so the correct answer is (A)

The specific heat capacity of water is:

0%
$1\ cal\ g^{-1}$
0%
$10\ cal\ g^{-1}$
0%
$2\ cal\ g^{-1}$
0%
$30\ cal\ g^{-1}$

Explanation

The specific heat capacity of water is

$1$ calorie g

$^{-1} \ ^oC ^{-1}$ or

$4.186$ joule

$g$

$^{-1} \ ^oC ^{-1}$ .
It means to increase the temperature of

$1$ g of water by

$1$ deg

$C$ , the heat required is

$1$ calorie or

$4.186$ J.

Which of the following parameters does not characterize the thermodynamic state of matter?

0%
temperature
0%
pressure
0%
work
0%
volume

Metabolism is the stepwise breakdown of the food we eat to provide energy for growth
and function. A general overall equation for this complex process represents the
degradation of glucose

$(C_{6}H_{12}O_{6})$ to

$CO_{2}$ and

$H_{2}O$ . This metabolic process involves many
steps and its enthalpy

$(\Delta H)$ is called the enthalpy of combustion. This is because the same
quantity of heat is evolved whether we burn 1 mole of glucose in air or let the metabolic
process break it down. Which of the following equations can be used to calculate the
standard enthalpy of the metabolic process correctly?

0%
$H^{o} = [\Delta_{f}H^{o} (CO_{2})+ \Delta_{f}H^{o} (H_{2}O)] [\Delta_{f}H^{o} (C_{6}H_{12}O_{6}) + \Delta_{f}H^{o} (O_{2})]$
0%
$H^{o} = [3\Delta_{f}H^{o} (CO_{2})+ 3\Delta_{f}H^{o} (H_{2}O)] [\Delta_{f}H^{o} (C_{6}H_{12}O_{6}) + 3\Delta_{f}H^{o} (O_{2})]$
0%
$H^{o} = [3\Delta_{f}H^{o} (CO_{2})+ 6\Delta_{f}H^{o} (H_{2}O)] [\Delta_{f}H^{o} (C_{6}H_{12}O_{6}) + 3\Delta_{f}H^{o} (O_{2})]$
0%
$H^{o} = [6\Delta_{f}H^{o} (CO_{2})+ 6\Delta_{f}H^{o} (H_{2}O)] [\Delta_{f}H^{o} (C_{6}H_{12}O_{6}) + 6\Delta_{f}H^{o} (O_{2})]$

Explanation

$(D) H^o=[6\Delta_fH^o(CO_2)+6\Delta_fH^o(H_2O)][\Delta_fH^o(C_6H_{12}O_6)+6\Delta_fH^o(O_2)]$

Because

$6$ molecules of

$CO_2$ along with

$6$ molecules of

$H_2O$ with

$1$ molecule of glucose and

$6$ molecules of

$H_2O$ is the standard enthalpy process equation we can take for the enthalpy calculation.

$C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O$ .

Option D is correct.

The total heat content of a system is

0%
Entropy
0%
Free energy
0%
Enthalpy
0%
Kinetic energy

$n \ moles$ of gas an ideal monatomic gas is confined in a cylinder A of volume

$V_{0}$ , pressure

$P_{0}$ and temperature

$T_{0}$ , this cylinder is connected to another cylinder B with the help of tube of a negligible volume. The cylinder B is fitted with a movable piston which can be adjusted from outside. Initially, the piston is adjusted so that volume of B is 7

$V_{0}$ and B is evacuated and then stopcork is opened so that gas expands and occupies the volume

$8V_{0}$ . [System is thermally isolated from the surroundings].

During this free expansion, the internal energy of the system

0%
Increases
0%
Decreases
0%
Remains constant
0%
Nothing can be said

A sample of oxygen gas expands its volume from

$3\ L\ to\ 5\ L$ against a constant pressure of

$3\ atm$ . If work done during expansion is used to heat 10 mole of water initially present at

$290\ K$ , its final temperature will be (specific heat capacity of water = 4.18 J/kg):

0%
$292.0\ K$
0%
$298.0\ K$
0%
$290.8\ K$
0%
$293.7\ K$

Explanation

Work done during expansion is equal to the heat supplied to water to raise the temperature.

$P({V}_{2} - {V}_{1}) = mS\Delta T$

$3\times 2\times 101.3 = 180\times 4.18 ( {T}_{f} - 290)$

${T}_{f} = 290.8 K$

Water of mass

$m_2$ = 1 kg is contained in a copper calorimeter of mass

$m_1$ = 1 kg. Their common temperature t =

$10^{0}C$ . Now a piece of ice of mass

$m_3$ = 2 kg and temperature is

$-11^{0}C$ dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper = 0.1 Kcal/ kg

$^{0}C$ , specific heat of water = 1 Kcal/kg

$^{0}C$ , specific heat of ice = 0.5 Kcal/kg

$^{0}C$ , latent heat of fusion of ice = 78.7 Kcal/kg]

0%
$0^{0}C$
0%
$4^{0}C$
0%
$- 4^{0}C$
0%
$- 2^{0}C$

Explanation

Loss in heat from calorimeter + water as temperture changes from

$10^{0}C$ to 0

$^{0}C$

$= m_1C_110 + m_2C_210 =$

$1 1 10 + 1 0.1 10 = 11 kca$ l
Gain in heat of ice as its temperature changes from

$- 11^{0}C$ to 0

$^{0}C$ $ kcal$
Hence ice and water will coexist at 0

$^{0}C$ without any phase change

Specific heat of a gas undergoing adiabatic change is

0%
1
0%
0
0%
4
0%
6

Explanation

During adiabatic process, no heat is supplied to system. The term

$\partial H$ is zero.Hence the specific heat

$\displaystyle \left ( \frac{\partial H}{\partial T} \right )$ is zero.

The energy stored in water formed from ice is:

0%
latent heat of fusion of ice
0%
$80\ cal\ g^{-1}$
0%
latent heat of boiling
0%
latent heat of fusion of water

Explanation

The energy stored in water formed from ice is latent heat of fusion of ice. It is 80 cal g

$^{-1}$ or 334 joules g

$^{-1}$ . It is the enthalpy change associated with melting of 1 g of ice to water.

Specific heat may be defined as:

0%
heat capacity at constant volume
0%
heat capacity at constant pressure
0%
heat capacity mol $^{-1}$
0%
heat capacity g $^{-1}$

The temperature at which the reaction

$AgO(s) \rightarrow Ag(s) + \frac{1}{2} O_2(g)$ is at equilibrium is?
Given

$\Delta H = 30.5 KJ mol^{-1}$ and

$\Delta S = 0.066 KJK^{-1} mol^{-1}$

0%
462.12 K
0%
362.12 k
0%
262.12 k
0%
562.12 k

Explanation

The temperature at which the reaction is at equilibrium is

$T=\frac {\Delta H}{Delta S}$ .
Substitute values in the above expression.

$T= \frac {\Delta H = 30.5 KJ mol^{-1}} {\Delta S = 0.066 KJK^{-1} mol^{-1}}=462.12K$

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

0%
$\Delta_rS$ for $\frac {1}{2} Cl_2(g)\rightarrow Cl(g)$ is positive.
0%
$\Delta E < 0$ for combustion of $CH_4(g)$ in a sealed container with rigid adiabatic system.
0%
$\Delta G$ is always zero for a reversible process in a closed system.
0%
$\Delta G^o$ for an ideal gas reaction is a function of pressure.

Explanation

$Cl_{2(g)}\rightarrow 2Cl_{(g)}$ Randomness is increasing.
(B) In closed container

$\Delta V=0$ hence work done is zero. There is no heat exchange.

$\therefore$

$\Delta E=q+W=0$
(C)

$\Delta G$ will be zero only when equilibrium is reached.
(D)

$\Delta G^o=-RT ln K_{eq}\therefore$ Not a function of pressure.

Actual flame temperature is always lower than the adiabatic flame temperature, because there is ______________.

0%
no possibility of obtaining complete combustion at high temperature.
0%
always loss of heat from the flame.
0%
both (a) and (b).
0%
neither (a) nor (b).

The internal energy of a compressed real gas, as compared to that of the normal gas at the same temperature, is

0%
less
0%
more
0%
sometimes less, sometimes more
0%
none of these

Two bodies at different temperatures are mixed in a calorimeter. Which of the following quantities remains conserved?

0%
sum of the temperatures of the two bodies
0%
total heat of the two bodies
0%
total internal energy of the two bodies
0%
internal energy of each body

Explanation

There is no heat lost since the calorimeter is insulated .
If we consider both the liquids together as one system , then the work done by the system is zero.

From first law of thermodynamics,

$\Delta U = Q - W$

$\Delta U = 0$

Since there is no change in internal energy, the internal energy of the system remains constant.

Hess's law is applicable for the determination of heat of:

0%
reaction
0%
transition
0%
formation
0%
all of the above

Which specific process has negative value of specific heat?

0%
Saturated vapours
0%
Ice
0%
Water
0%
Vapours

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%
none of these

A liquid boils at such a temperature at which the saturated vapour pressure, as compared to atmospheric pressure, is

0%
one-third
0%
equal
0%
half
0%
double

The internal energy of a gram-molecule of an ideal gas depends on

0%
pressure alone
0%
volume alone
0%
temperature alone
0%
both pressure as well as temperature

Explanation

The internal energy of an ideal gas is given by the expression :

$U = C_{v}T$ .
From the above expression its clear that internal energy depend only on temperature.

6.80 g of

$NH_3$ ; is passed over heated CuO. The standard heat enthalpies of

$NH_{3}$ ,

$CuO(s)$ and

$H_2O(l)$ are -46.0, -155.0 and

$-285 0 \: kJ \: mol^{-1}$ respectively and the change is.

$NH_3 +(3/2)CuO\longrightarrow \frac{1}{2}N_2(g)+ (3/2)H_2O(l) +(3/2)Cu(s)$ .
If the enthalpy change is

0%
-50
0%
-59.6
0%
60
0%
30

Explanation

$NH_3 + (3/2)CuO\longrightarrow \frac{1}{2}N_2(g)+ (3/2)H2O(l)+ (3/2)Cu(s) ; \: \Delta H=?$

$\Delta H^{\circ}=\Sigma H^{\circ}_{Products}-\Sigma H^{\circ}_{Reactants}$

$\displaystyle =[\frac{1}{2}\times H^{\circ}_{N_2}+( 3 / 2)\times H^{\circ}_{H_2O(l)}+~(3 /2) H^{\circ}_{Cu}]-[(3/2)H^{\circ}_{CuO}+H^{\circ}_{NH_3}]$

$=[0+ (3/2)\times (-285)+ (3/2)\times 0]-[(3/2) \times(-155)+ (-46)]$

$=-149\,kJ$

$[\because H^{\circ} for \: pure \: element = 0 \: and \: H^{\circ}_{compound} =H^{\circ}_{formation}]$

$\because 17 \: g \: NH_3 \: brings \: in \: change \: in \: \Delta H = 149 \: kJ$

$\because 6.8 \: g \: NH_3 \: brings \: in \: change \: in \: \Delta H = (149 \times 6.8)/17 = 59.6 \,kJ$

$\therefore \Delta H for \: 6.89 \: g \: NH_3 = - 59.6 \: kJ$

The enthalpy change for the reaction is:

$XeF_4\longrightarrow Xe^+ + F^- + F_2 + F; \: \ \ \ \ \Delta H=?$

The average

$Xe-\! \! \! -F$ bond energy is

$34 \:kcal \:mol^{-1}$ .

${IE}_1$ of Xe is

$279 \: kcal \: mol^{-1}$ and electron affinity of F is

$85 \:kcal \:mol^{-1}$ and

$e_{F-\! \! \! -F} = 38 \:kcal \: mol^{-1}$ .

0%
$345 \:kcal \: mol^{-1}$
0%
$292 \:kcal \: mol^{-1}$
0%
$174 \:kcal \: mol^{-1}$
0%
None of these

Explanation

Bond energy

$={ BE }_{ { X }_{ e }-F }=34kCal/mol$
So,

${ X }_{ e }{ F }_{ 4 }=4 ({ X }_{ e }-F)$ bonds

$\Rightarrow$ total bond energy

$=4\times 34kCal/mol \Rightarrow{ BE }_{ { X }_{ e }-{ F }_{ 4 } }=136kCal/mol$
Now this amount of bond energy is liberated during formation of

${ X }_{ e }{ F }_{ 4 }$

$\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }=-136kCal/mol$

${ X }_{ e }+4F\rightarrow { X }_{ e }{ F }_{ 4 }$

$\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }=-136kCal/mol$ ...(1)
Now

$I{ E }_{ 1 }$ of

${ X }_{ e }=279kCal/mol$
So

${ X }_{ e }\rightarrow { { X }_{ { e } } }^{ + }+{ e }^{ - }$

$\Delta H=+I{ E }_{ 1 }$ ...(2)
Now,

$EA$ of

$F=85kCal/mol$

$F\rightarrow { F }^{ - }-{ e }^{ - }$

$\Delta H=-{ E }A$ ...(3)
(energy is released in this process)
and

$2F\rightarrow{ F }_{ 2 }$

$\Delta K=-{ e }_{ F-F }$ ...(4)
Now

$(1)$ becomes

$\Rightarrow { X }_{ e }{ F }_{ 4 }\rightarrow { X }_{ e }+4F$

$\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }=136kCal/mol$ ...(5)
Now,

$(2)+(3)+(4)+(5)$ gives

${ X }_{ e }{ F }_{ 4 }+{ X }_{ e }+3F\rightarrow { X }_{ e }+4F+{ X }_{ e }^{ + }+{ e }^{ - }+{ F }^{ - }-{ e }^{ - }+{ F }_{ 2 }$

$\Delta H=\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }+\left( H{ E }_{ 1 } \right) +\left( -EA \right) +\left( -{ e }_{ F-F } \right) =136+279+(-85)+(-38)=292$

$\Rightarrow { X }_{ e }{ F }_{ 4 }={ X }_{ e }^{ + }+{ F }^{ - }+{ F }_{ 2 }+F$

$\Delta H=292kCal/mol$

The internal energy of a piece of lead when beaten by a hammer will

0%
increase
0%
decrease
0%
remain constant
0%
sometimes increases and sometimes decreases

For a given reaction,

$\Delta H=35.5\ kJ\ mol^{-1}$ and $$ \Delta S=83.6\ J\ K^{-1}\ mol^{-1}. The reaction is spontaneous at:

(Assume that

$\Delta H$ and

$\Delta S$ so not vary with temperature)

0%

$T<425K$
0%
$T>425K$
0%
all temperatures
0%
$T>298K$

Hess's law is related to:

0%
equilibrium constant
0%
change in heat during a reaction
0%
rates of reaction
0%
influence of pressure on volume of a gas

The heat capacities of the following gases at room temperature are such that :

0%
$NH_3 > CO_2 = O_2 = N_2 > Ar$
0%
$NH_3 = CO_2 > O_2 > N_2 > Ar$
0%
$NH_3 > CO_2 > O_2 = N_2 > Ar$
0%
$NH_3 > CO_2 > O_2 > N_2 > Ar$

Explanation

Heat capacity of gases increase at room temperature by increasing the atomicity.

Order of the heat capacity is

$NH_3 > CO_2 > O_2 = N_2 > Ar$
4 3 2 2 1

Hence,option C is correct.

What is the free energy change

$(\Delta G)$ when 1.0 mole of water at

$100^{\circ}\! C$ and 1 atm pressure is converted into steam at

$100^{\circ}\! C$ and 1 atm pressure?

0%
80 cal
0%
540 cal
0%
620 cal
0%
zero

Explanation

$H_2O_{(l)} \longrightarrow H_2O_{(g)}$
373 K 373 K
1atm 1atm

$\displaystyle \Delta S=\frac{\Delta H_{vap}}{T}$

$\displaystyle \Delta G=\Delta H_f - \Delta H_i=0$

Hence, the correct option is

$D$

Chemical reaction is invariably associated with the transfer of energy either in the form of heat or light. In the laboratory, heat changes in physical and chemical processes are measured with an instrument called Calorimeter. Heat change in the process is calculated as :

$q = ms\Delta T$ ........ $s= specific\quad heat$
$= c\Delta T$ ....... $c= heat\quad capacity$

Heat of reaction at constant volume is measured using bomb Calorimeter.
$q_V = \Delta U =$ internal energy change

Heat of reaction at constant pressure is measured using bomb Calorimeter.
$q_P = \Delta H$
$q_P = q_V + P\Delta V$
$\Delta H = \Delta U + \Delta nRT$

The heat capacity of a bomb calorimeter is

$500 J K^{ -1 }$ . When

$0.1 g$ of Methane was burnt in this calorimeter, the temperature rose by

$2^{ \circ }C$ . The value of

$\Delta U$ per mole will be :

0%
$-160 kJ$
0%
$+260 kJ$
0%
$-2 kJ$
0%
$+2 kJ$

Explanation

Internal energy of the system can be defined as :

$\Delta E=n{ C }_{ v }\Delta T$

$n=\cfrac { m }{ M } =\cfrac { 0.1 }{ 16 } =6.25\times{ 10 }^{ -3 }mol$

Specific heat of methane is

${ C }_{ v }=500KJ{ mol }^{ -1 }$

${ T }_{ 1 }=273K\\ { T }_{ 2 }=273+2=275K$

$\Delta E=n{ C }_{ v }\Delta T$

$\Delta E=6.25\times{ 10 }^{ -3 }\times500\times(275-273)=-160KJ/mol\quad \quad$

Therefore, internal energy per mole will be -160KJ/mol.

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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