Thermodynamics Test

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Thermodynamics Test - 37

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Question 1
1 / -0

Which heat is produced throughout the conducting wire?

A
Petlier heat

B
Thomson effect heat

C
Joule heat

D
none of these

Solution

As the heat produced in the conductor by following Joule's law of heating or cooling. Sp, the produced heat is Joule heat.
Question 2
1 / -0

The value of $$\triangle $$H and $$\triangle $$S for the reaction, $${ C }_{ \left( graphite \right) }+{ CO }_{ 2\left( g \right) }\rightarrow 2{ CO }_{ (g) }$$ are 170kJ and $${ 170JK }^{ -1 }$$, respectively. This reaction will be spontaneous at:

A
510K

B
710K

C
910K

D
1119K

Solution

$$C_{(graphite)} + CO_{2}(g) \rightarrow 2CO(g)$$
The $$Rn$$ is spontaneous when $$ \Delta G \leqslant O $$
$$ \Rightarrow \Delta H - T\Delta S \leq O $$
$$ \Rightarrow T\Delta S \geqslant \Delta H$$
$$ \Rightarrow T\geqslant \dfrac{\Delta H}{\Delta S} $$
$$ T \geqslant \dfrac{170 \times 10^{3}}{1170}$$
$$ T \geqslant 1000K $$
Question 3
1 / -0

For an exothermic reaction to be sponteneous:

A
temperature must be high

B
temperature must be zero

C
temperature may have any magnitude

D
temperature and $$\Delta H < T \Delta S$$.

Solution

According to second law of thermodynamics,
$$\Delta G\quad =\quad \Delta H-T\Delta S$$.
For a reaction to be spontaneous, $$\Delta G$$ have to be $$-ive$$.
For, an exothermic reaction, $$\Delta H$$ is always $$-ve$$, which makes the value of $$\Delta G$$ $$-ve$$ at any temperature.
So, exothermic reaction is spontaneous at any temperature.
So, correct answer is option C.
Question 4
1 / -0

For a reversible spontaneous change $$\triangle s$$ is:

A
$$\dfrac { \triangle E }{ T } $$

B
$$\dfrac { P\triangle V }{ T } $$

C
$$\dfrac { q }{ T } $$

D
$$RTlogK$$

Solution

$$ds=\dfrac{dq.rev}{T}$$
$$\displaystyle \triangle s =\dfrac{\int dq.rev}{T}$$
$$\displaystyle \int dq.rev.q$$ (for rev. provess)
$$\triangle s=\dfrac{q}{T}$$
Question 5
1 / -0

For a process to be spontaneous at constant $$T$$ and $$P$$:

A
$$(\Delta G)_{system}$$ must be negative

B
$$(\Delta G)_{system}$$ must be positive

C
$$(\Delta S)_{system}$$ must be positive

D
$$(\Delta S)_{system}$$ must be negative

Solution

The standard free energy change by using the standard entropies and enthalpies of a reaction.
When $$\Delta G=-ve$$, the change in the enthalpy and entropy will be positive and which is depends on the temperature and spontaneity ocuurs due to high temperature.
Question 6
1 / -0

Which of the following must be spontaneous at all temperatures?

A
$$\Delta H_{sys}=-ve; \Delta S_{sys}=-ve$$

B
$$\Delta H_{sys}=-ve; \Delta S_{sys}=+ve$$

C
$$\Delta H_{sys}=+ve; \Delta S_{sys}=+ve$$

D
$$\Delta H_{sys}=+ve:; \Delta S_{sys}=-ve$$

Solution

A reaction is spontaneous when the Gibbs free energy change ($$\Delta{G}<0$$)

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

where $$\Delta{H}$$ is the change in enthalpy and $$\Delta{s}$$ is the change in entropy.

A. $$\Delta H=-Ve$$ $$\Delta S$$ = $$-ve$$ at low temperatures reaction is spontaneous.

B. $$\Delta H=-Ve$$ $$\Delta S$$ = $$+ve$$ at all temperatures reaction is spontaneous.

C. $$\Delta H=+Ve$$ $$\Delta S$$ = $$+ve$$ at high temperatures reaction is spontaneous.

D. $$\Delta H=+Ve$$ $$\Delta S$$ = $$-ve$$ at all temperatures reaction is non-spontaneous.

Note: Low $$T$$ and high $$T$$ is relative to the entropy. Here, low $$T$$ means the $$T-factor$$ can not dominate $$\Delta{S}-factor$$. However, high $$T$$ means that the $$T-factor$$ will dominate $$\Delta{S}-factor$$.

Hence, the correct option is B.
Question 7
1 / -0

Specific heats of monoatomic and diatomic gases are same and
satisfy the relation which is

A
$${C_p}\left( {mono} \right) = {C_p}\left( {dia} \right)$$

B
$${C_p}\left( {mono} \right) = {C_\upsilon }\left( {dia} \right)$$

C
$${C_\upsilon }\left( {mono} \right) = {C_\upsilon }\left( {dia} \right)$$

D
$${C_\upsilon }\left( {mono} \right) = {C_p}\left( {dia} \right)$$

Solution
Question 8
1 / -0

$$Pt$$ | $$Cl_2$$ ($$P_1$$ atm) | $$HCl$$ ($$0.1$$M) | $$Cl_2$$ ($$P_2$$ atm) | $$Pt$$, cell reaction will be spontaneous if:

A
$$P_1 = P_2$$

B
$$P_1 > P_2$$

C
$$P_2 > P_1$$

D
$$P_1 = P_2 =1$$ atm

Solution

Solution:- (C) $${P}_{2} > {P}_{1}$$
$${E}_{cell} = {{E}^{0}}_{cell} + \cfrac{0.059}{2} \log{\cfrac{{P}_{2}}{{P}_{1}}}$$
For cell reaction to be spontaneous,
$${E}_{cell} > 0$$
$$\therefore {E}_{cell}$$ will be $$+ve$$ if $${P}_{2} > {P}_{1}$$.
Hence for the cell reaction to be spontaneous,
$${P}_{2} > {P}_{1}$$
Question 9
1 / -0

In the series $$Sc(Z=21)$$ to $$Zn(Z=30)$$, the enthalpy of atomisation of which element is least?

A
Sc

B
Mn

C
Cu

D
Zn

Solution

$$Sc\ \& \ Zn$$ belongs to the third group of the periodic table. The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals, there are some unpaired electrons that account for their stronger metallic bonding.
Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in $$Zn$$ and as a result, it has the least enthalpy of atomization.
Question 10
1 / -0

For a first order reaction rate constant is $$1 \times {10^{ - 5}}{\sec ^{ - 1}}$$ having $${E_a} = 1800\ kJ/mol$$ . Then the value of $$\ell nA$$ at$$T = 600\ K$$ is:-

A
$$151.7$$

B
$$349.3$$

C
$$24.7$$

D
$$11.34$$

Solution

acc. to arrhenius equation
$$k=A{ e }^{ -Ea/RT }\quad $$
rate constant $$1\times { 10 }^{ -5 }{ sec }^{ -1 }$$
$$Ea=1800kJ/mol$$
$$\log { k } =\log { A } -\cfrac { Ea }{ RT } $$
$$\log { \left( 1\times { 10 }^{ -5 } \right) } =\log { A } -\cfrac { Ea }{ RT } $$
$$\log { A } =\log { \left( 1\times { 10 }^{ -5 } \right) } +\cfrac { 1800 }{ 1\times { 10 }^{ -5 }\times 600 } =151.7$$

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Thermodynamics Test - 37

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Thermodynamics Test - 37

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SHARING IS CARING

Question 1

Petlier heat

Thomson effect heat

Joule heat

none of these

Solution

Question 2

510K

710K

910K

1119K

Solution

Question 3

temperature must be high

temperature must be zero

temperature may have any magnitude

temperature and $$\Delta H < T \Delta S$$.

Solution

Question 4

$$\dfrac { \triangle E }{ T } $$

$$\dfrac { P\triangle V }{ T } $$

$$\dfrac { q }{ T } $$

$$RTlogK$$

Solution

Question 5

$$(\Delta G)_{system}$$ must be negative

$$(\Delta G)_{system}$$ must be positive

$$(\Delta S)_{system}$$ must be positive

$$(\Delta S)_{system}$$ must be negative

Solution

Question 6

$$\Delta H_{sys}=-ve; \Delta S_{sys}=-ve$$

$$\Delta H_{sys}=-ve; \Delta S_{sys}=+ve$$

$$\Delta H_{sys}=+ve; \Delta S_{sys}=+ve$$

$$\Delta H_{sys}=+ve:; \Delta S_{sys}=-ve$$

Solution

Question 7

$${C_p}\left( {mono} \right) = {C_p}\left( {dia} \right)$$

$${C_p}\left( {mono} \right) = {C_\upsilon }\left( {dia} \right)$$

$${C_\upsilon }\left( {mono} \right) = {C_\upsilon }\left( {dia} \right)$$

$${C_\upsilon }\left( {mono} \right) = {C_p}\left( {dia} \right)$$

Solution

Question 8

$$P_1 = P_2$$

$$P_1 > P_2$$

$$P_2 > P_1$$

$$P_1 = P_2 =1$$ atm

Solution

Question 9

Sc

Mn

Cu

Zn

Solution

Question 10

$$151.7$$

$$349.3$$

$$24.7$$

$$11.34$$

Solution

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