Thermodynamics Test

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Thermodynamics Test - 29

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Question 1
1 / -0

The value of $$\Delta H$$ and $$\Delta S$$ for a reaction are respectively $$30kJ$$ $${mol}^{-1}$$ and $$100J{K}^{-1}$$ $${mol}^{-1}$$. Then temperature above which the reaction will become spontaneous is:

A
$$300K$$

B
$$30K$$

C
$$100K$$

D
$${300}^{o}C$$

Solution

Gibb's free energy equation : $$\Delta G=\Delta H-T\Delta S$$
For reaction to be spontaneous, $$\Delta G<0$$
$$\therefore \quad \Delta H-T\Delta S<0$$
$$\therefore \quad \left( 30\times { 10 }^{ 3 } \right) -T\times \left( 100 \right) <0$$
$$\therefore \quad 30000<100T$$
$$\therefore \quad 300<T$$
$$\therefore \quad T>300K$$
Therefore, for reaction to be spontaneous, temperature should be above 300 K.
Question 2
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Which of the following thermodynamics relation is correct?

A
$$dG=VdP-SdT$$

B
$$dU=PdV+TdS$$

C
$$dH=-Vdp+TdS$$

D
$$dG=VdP+SdT$$

Solution

$$dG=dH-TdS-SdT$$ ... $$(G=H-TS)$$
$$dH=dU+PdV+VdP$$ ... $$(H=U+PV)$$
and $$dU=TdS-PdV$$
$$\therefore$$ $$dG=(TdS-PdV)+PdV+VdP-TdS-SdT$$
$$dG=VdP-SdT$$
Question 3
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$$\Delta H$$ and$$\Delta S$$ for the reaction
$${ Br }_{ 2 }(l)+{ Cl }_{ 2 }(l)\longrightarrow 2BrCl(g)$$
are $$29.37\ kJ$$ and $$104.0\ J{ K }^{ -1 }$$ respectively. Above that temperature will this reaction become spontaneous?

A
T > 177.8 K

B
T > 35.1 K

C
T > 28.4 K

D
T > 141.2 K

Solution

According to Gibbs-Helmholtz equation
$$\Delta G=\Delta H-T\Delta S$$
For spontaneous process, $$\Delta G< 0$$
i.e., $$T\Delta S>\Delta H$$
$$T>\cfrac { \Delta H }{ \Delta S } $$
$$T>\cfrac { 29.37\times 1000 }{ 104 } $$
$$T> 282.4K$$
Question 4
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A $$1g$$ sample of substance $$A$$ at $${100}^{o}C$$ is added to $$100mL$$ of $${H}_{2}O$$ at $${25}^{o}C$$. Using separate $$100mL$$ portion of $${H}_{2}O$$ the procedure is repeated with substance $$B$$ then with substance $$C$$. How will the final temperatures of the water compare?
Substance Specific heat
$$A$$ $$0.6{g}^{-1}$$ $$^{ o }{ { C }^{ -1 } }$$
$$B$$ $$0.4{g}^{-1}$$ $$^{ o }{ { C }^{ -1 } }$$
$$C$$ $$0.2{g}^{-1}$$ $$^{ o }{ { C }^{ -1 } }$$

A
$${T}_{C}>{T}_{B}> {T}_{A}$$

B
$${T}_{B}>{T}_{A}> {T}_{C}$$

C
$${T}_{A}>{T}_{B}> {T}_{C}$$

D
$${T}_{A}= {T}_{B}= {T}_{C}$$

Solution

$$\text{More the value of Specific heat more the value of final temperature.}$$
$$T_A>T_B>T_C$$
Question 5
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The Haber's process for production of ammonia involves the equilibrium:
$${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons 2{NH}_{3}(g)$$
Assuming $$\Delta {H}^{o}$$ and $$\Delta {S}^{o}$$ for the reaction do not change with temperature, which of the statements is true?
($$\Delta {H}^{o}=-95kJ$$ and $$\Delta {S}^{o}=-190J{ K }^{ -1 }$$)

A
Ammonia dissociates spontaneously below $$500K$$

B
Ammonia dissociates spontaneously above $$500K$$

C
Ammonia dissociates at all temperatures

D
Ammonia does not dissociate at any temperature

Solution

$$\Delta { G }^{ o }=\Delta { H }^{ o }-T\Delta { S }^{ o }$$
$$=-95\times 1000-500\times (-90)$$
$$=0$$
Thus at temperatures below 500K, $$\Delta G <0$$.
Question 6
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For a particular reaction, $$\Delta { H }^{ o }=-38.3\ kJ$$ and $$\Delta { S }^{ o }=-113\ J{ K }^{ -1 }{ mol }^{ -1 }$$. This reaction is:

A
spontaneous at all temperatures

B
non-spontaneous at all temperatures

C
spontaneous at temperatures below $${66}^{o}C$$

D
spontaneous at temperatures above $${66}^{o}C$$

Solution

$${ \left( \Delta { G } \right) }=$$$${ \left( \Delta { H } \right) }-T$$$${ \left( \Delta { S } \right) }$$
for spontanious $${ \left( \Delta { G } \right) }$$ must be negative.
$${ \left( \Delta { H } \right) }-T$$$${ \left( \Delta { S } \right) }$$<0
$${ \left( \Delta { H } \right) }<T$$$${ \left( \Delta { S } \right) }$$
$${ \left( \Delta { H } \right) }$$/$${ \left( \Delta { S } \right) }$$<$$T$$
$$(-38300)$$/$$(-113)$$<$$T$$
$$339K$$<$$T$$
$$66^oC$$<$$T$$
Question 7
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Following enthalpy changes are given:

$$\alpha -D\quad glucose(s)\longrightarrow \alpha -D\quad glucose(aq.);\quad \Delta H=10.72kJ$$

$$\beta -D\quad glucose(s)\longrightarrow \beta -D\quad glucose(aq.);\quad \Delta H=4.68kJ$$

$$\alpha -D\quad glucose(aq.)\longrightarrow \beta -D\quad glucose(aq.);\quad \Delta H=1.16kJ$$

Calculate the enthalpy change in,

$$\alpha -D\quad glucose(s)\longrightarrow \beta -D\quad glucose(s)$$

A
$$14.24kJ$$

B
$$16.56kJ$$

C
$$7.2kJ$$

D
$$4.88kJ$$

Solution
Question 8
1 / -0

For which process will $$\Delta { H }^{ o }$$ and $$\Delta { G }^{ o }$$ be expected to be most similar?

A
$$2Al(s)+{ Fe }_{ 2 }{ O }_{ 3 }(s)\longrightarrow 2Fe(s)+{ Al }_{ 2 }{ O }_{ 3 }(s)$$

B
$$2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq.)+{ H }_{ 2 }(g)$$

C
$$2N{ O }_{ 2 }(g)\longrightarrow { N }_{ 2 }{ O }_{ 4 }(g)$$

D
$$2{ H }_{ 2 }(g)+{ O }_{ 2 }(g)\longrightarrow 2{ H }_{ 2 }O(g)$$

Solution
Question 9
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Consider the values of $$\Delta { H }^{ o }$$ (in $$kJ{ mol }^{ -1 }$$) and for $$\Delta { S }^{ o }$$ (in $$J{ mol }^{ -1 }{ K }^{ -1 }$$) given for four different reactions. For which reaction will $$\Delta { G }^{ o }$$ increase the most positive) when temperature is increased from $${0}^{o}C$$ to $${25}^{o}C$$?

A
$$\Delta { H }^{ o }=50,\Delta { S }^{ o }=50\quad $$

B
$$\Delta { H }^{ o }=90,\Delta { S }^{ o }=20\quad $$

C
$$\Delta { H }^{ o }=-20,\Delta { S }^{ o }=-50\quad $$

D
$$\Delta { H }^{ o }=-90,\Delta { S }^{ o }=-20\quad $$

Solution

$${ \left( \Delta { G } \right) }=$$$${ \left( \Delta { H } \right) }-T$$$${ \left( \Delta { S } \right) }$$
$$\text{from the above equation we can easily clearify that more the negative value of }$$ $${ \left( \Delta { S } \right) }$$ $$\text{increases most positive the value of }$$$${ \left( \Delta { G } \right) }$$.
$$\text{So option C is correct.}$$
Question 10
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The standard Gibbs free energy $$\Delta { G }^{ o }$$ is related to equilibrium constant $${K}_{P}$$ as:

A
$${K}_{P}=-RT\log{\Delta { G }^{ o }}$$

B
$${K}_{P}={[e/RT]}^{\Delta { G }^{ o }}$$

C
$${K}_{P}=-\Delta { G }^{ o }/RT$$

D
$${K}_{P}={e}^{-\Delta { G }^{ o }/RT}$$

Solution

For a reaction, the Gibb's free energy equation is :
$$\Delta G=\Delta { G }^{ 0 }+RTln{ K }_{ P }$$
At equilibrium, $$\Delta G=0$$
$$\therefore \quad { \Delta G }^{ 0 }=-RTln{ K }_{ P }$$
$$\therefore \quad { K }_{ P }={ e }^{ -{ \Delta G }^{ 0 }/RT }$$

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Substance	Specific heat
$$A$$	$$0.6{g}^{-1}$$ $$^{ o }{ { C }^{ -1 } }$$
$$B$$	$$0.4{g}^{-1}$$ $$^{ o }{ { C }^{ -1 } }$$
$$C$$	$$0.2{g}^{-1}$$ $$^{ o }{ { C }^{ -1 } }$$

Thermodynamics Test - 29

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Thermodynamics Test - 29

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Question 1

$$300K$$

$$30K$$

$$100K$$

$${300}^{o}C$$

Solution

Question 2

$$dG=VdP-SdT$$

$$dU=PdV+TdS$$

$$dH=-Vdp+TdS$$

$$dG=VdP+SdT$$

Solution

Question 3

T > 177.8 K

T > 35.1 K

T > 28.4 K

T > 141.2 K

Solution

Question 4

$${T}_{C}>{T}_{B}> {T}_{A}$$

$${T}_{B}>{T}_{A}> {T}_{C}$$

$${T}_{A}>{T}_{B}> {T}_{C}$$

$${T}_{A}= {T}_{B}= {T}_{C}$$

Solution

Question 5

Ammonia dissociates spontaneously below $$500K$$

Ammonia dissociates spontaneously above $$500K$$

Ammonia dissociates at all temperatures

Ammonia does not dissociate at any temperature

Solution

Question 6

spontaneous at all temperatures

non-spontaneous at all temperatures

spontaneous at temperatures below $${66}^{o}C$$

spontaneous at temperatures above $${66}^{o}C$$

Solution

Question 7

$$14.24kJ$$

$$16.56kJ$$

$$7.2kJ$$

$$4.88kJ$$

Solution

Question 8

$$2Al(s)+{ Fe }_{ 2 }{ O }_{ 3 }(s)\longrightarrow 2Fe(s)+{ Al }_{ 2 }{ O }_{ 3 }(s)$$

$$2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq.)+{ H }_{ 2 }(g)$$

$$2N{ O }_{ 2 }(g)\longrightarrow { N }_{ 2 }{ O }_{ 4 }(g)$$

$$2{ H }_{ 2 }(g)+{ O }_{ 2 }(g)\longrightarrow 2{ H }_{ 2 }O(g)$$

Solution

Question 9

$$\Delta { H }^{ o }=50,\Delta { S }^{ o }=50\quad $$

$$\Delta { H }^{ o }=90,\Delta { S }^{ o }=20\quad $$

$$\Delta { H }^{ o }=-20,\Delta { S }^{ o }=-50\quad $$

$$\Delta { H }^{ o }=-90,\Delta { S }^{ o }=-20\quad $$

Solution

Question 10

$${K}_{P}=-RT\log{\Delta { G }^{ o }}$$

$${K}_{P}={[e/RT]}^{\Delta { G }^{ o }}$$

$${K}_{P}=-\Delta { G }^{ o }/RT$$

$${K}_{P}={e}^{-\Delta { G }^{ o }/RT}$$

Solution

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