Thermodynamics Test

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Thermodynamics Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

The three thermodynamic states $$P, Q$$ and $$R$$ of a system are connected by the paths shown in the figure given on the right. The entropy change in the processes $$P \rightarrow Q , Q \rightarrow R$$ and $$P\rightarrow R$$ along the paths indicated are $$\triangle S_{P Q}, \triangle S_{QR}$$ and $$\triangle S_{PR}$$ respectively. If the process $$P \rightarrow Q$$ is adiabatic and irreversible, while $$P\rightarrow R$$ is adiabatic and reversible, the correct statement is:

A
$$\triangle S_{QR} > 0$$

B
$$\triangle S_{PR} > 0$$

C
$$\triangle S_{QR} < 0$$

D
$$\triangle S_{PQ} > 0$$

Solution

It is given that $$P\rightarrow R$$ is an adiabatic and reversible process.
For reversible adiabatic process, $${q}_{reversible}=0$$
So $$\triangle S=\int { \cfrac { { dq }_{ reversible } }{ T } } =0$$
Hence $$\triangle {S}_{PR}=0$$
The entropy of irreversible process is always going to increase. So, in an adiabatic irreversible process, change of entropy due to internal irreversibility is greater than zero.
Hence $$\triangle {S}_{PQ}=\cfrac{\triangle q}{T}\gt 0$$
entropy change of a cyclic process is always zero.
Since entropy is a state function i.e. it doesn't depend on the path
and $$\triangle {S}_{PR}=0\\ \triangle {S}_{PQ}\rightarrow (+)ve$$
Then $$\triangle {S}_{QR}$$ must be $$-ve$$.
Hence $$\triangle {S}_{QR}\lt 0$$
Question 2
1 / -0

Given the following data:
Substrate $$\Delta H^o$$(kJ/mol) $$S^o$$(J/mol K) $$\Delta G^o$$ (kJ/mol)
FeO(s) $$-266.3$$ $$57.49$$ $$245.12$$
C(Graphite) $$0$$ $$5.74$$ $$0$$
Fe(s) $$0$$ $$27.28$$ $$0$$
CO(g) $$-110.5$$ $$197.6$$ $$-137.15$$
Determine at what temperature the following reaction spontaneous?
$$FeO(s)+C(Graphite)\rightarrow Fe(s)+CO(g)$$

A
$$298$$ K

B
$$668$$ K

C
$$966$$ K

D
$$\Delta G^o$$ is $$+$$ve, hence the reaction will never be spontaneous

Solution
Question 3
1 / -0

In which of the following pair of reactions first reaction is spontaneous while second reaction is non spontaneous?

A
(i) $$^{-}SH+H_2O\rightarrow H_2S+OH^-$$

(ii) $$NH^-_2+H_2O\rightarrow NH_3+OH^-$$

B
(i) $$^{-}OR+H_2SO_4\rightarrow ROH+HSO^-_4$$

(ii) $$R^-+NH_3\rightarrow RH+NH^-_2$$

C
(i) $$Cl^{-}+HF\rightarrow HCl+F^-$$

(ii) $$^-OH+HCl\rightarrow H_2O+Cl^-$$

D
(i) $$^-OH+HBr\rightarrow H_2O+Br^-$$

(ii) $$RO^-+NH_3\rightarrow ROH+NH^-_2$$

Solution

(A) (i) $$\underset {base} {^{\circleddash}SH}+\underset {acid}{H_2O} \longrightarrow H_2S+OH^{\circleddash}$$ (ii) $$\underset {base}{NH_2^{\circleddash}}+\underset {acid}{H_2O}\longrightarrow NH_3+OH^{\circleddash}$$
Here both reactions are spontaneous. As $$SH^{\circleddash}$$ and $$NH_2^{\circleddash}$$ are highly basic.

(B) (i) $$\underset {base}{^{\circleddash} OR} +\underset {acid}{H_2SO_4} \longrightarrow ROH+HSO_4^{\circleddash}$$ (ii) $$\underset {base}{R^{\circleddash}}+\underset {acid}{NH_3}\longrightarrow RH+NH_2^{\circleddash}$$
Both reactions are spontaneous $$R^{\circleddash}$$ and $$OR^{\circleddash}$$ are highly basic.

(C) (i) $$\underset {base}{Cl^-}+\underset {acid}{HF}\longrightarrow HCl+F^{\circleddash}$$ (ii) $$\underset {base}{^{\circleddash}OH}+\underset {acid}{HCl} \longrightarrow H_2O+Cl^{\circleddash}$$
Here also both reactions are spontaneous.

(D) (i) $$\underset {base}{OH^{\circleddash}}+\underset {acid}{HBr}\longrightarrow H_2O+Br^{\circleddash}$$ (ii) $$\underset {base}{RO^{\circleddash}}+\underset {acid}{NH_3} \nrightarrow ROH+NH_2^{\circleddash}$$
(i) reaction is spontaneous but (ii) is not spontaneous. Because basicity of $$RP^{\circleddash}$$ is not high enough to abstract proton from weak acid $$NH_3$$ .
Question 4
1 / -0

If $$\Delta H> 0$$ and $$\Delta S> 0$$, the reaction proceeds spontaneously when:

A
$$\Delta H > T \Delta S$$

B
$$\Delta H < T \Delta S$$

C
$$\Delta H=T\Delta S$$

D
$$None\ of\ the\ above$$
Question 5
1 / -0

Which of the following conditions regarding a chemical process ensures its spontanlity at all temperature?

A
$$\triangle H>0,\quad \triangle G<0$$

B
$$\triangle H<0,\quad \triangle S>0$$

C
$$\triangle H<0,\quad \triangle S<0$$

D
$$\triangle H>0,\quad \triangle S<0$$

Solution

$$\triangle H<0$$, $$\triangle S>0$$
$$\Rightarrow$$ Reaction spontaneous at all temperature.
$$\rightarrow$$ For spontaneity, $$\triangle G<0$$
$$\Rightarrow \triangle H- T \triangle S <0$$
where $$\triangle H$$ is negative and $$\triangle S$$ is positive.
Question 6
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When a block of iron floats in mercury at $$0^o$$C, a fraction $$k_1$$ of its volume is submerged, while at the temperature $$60^o$$C, a fraction $$k_2$$ is seen to be submerged. If the coefficient of volume expansion of iron is $$\gamma _{Fe}$$ and that of mercury is $$\gamma_{Hg}$$, then the ratio $$k_1/k_2$$ can be expressed as.

A
$$\dfrac{1+60\gamma_{Fe}}{1+60\gamma_{Hg}}$$

B
$$\dfrac{1-60\gamma_{Fe}}{1+60\gamma_{Hg}}$$

C
$$\dfrac{1+60\gamma_{Fe}}{1-60\gamma_{Hg}}$$

D
$$\dfrac{1+60\gamma_{Hg}}{1+60\gamma_{Fe}}$$

Solution
Question 7
1 / -0

Which of the following statements is correct for a reverse process in a state of equilibrium ?

A
$$\Delta G=2.30\quad RT\quad log\quad K$$

B
$$\Delta { G }^{ o }=-2.30\quad RT\quad log\quad K$$

C
$$\Delta { G }^{ o }=2.30\quad RT\quad log\quad K$$

D
$$\Delta { G }=-2.30\quad RT\quad log\quad K$$

Solution

The correct option is C
$$\Delta G^0=-2.303 RT log K$$
$$\Delta=\Delta G^0+2.30 RT log K log Q$$
At equilibrium when
$$\Delta=O and Q$$
$$\Delta G =\Delta G^0+2.303 RT log K=0$$

$$\Delta G^0=-2.303 RT log K$$
$$\Delta G=\Delta G^0+2.303 RT log KQ$$
At equilibrium when
$$\Delta G= \Delta G^0+2.303 RT log K=0$$
$$\Delta^0=-2.303 RTK$$
Question 8
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In a constant volume calorimeter, 5 g of gas with molecular weight 40 was burnt in excess of oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.75 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ$${ K }^{ -1 }$$, the numerical value for the $$\triangle U$$ of combustion of the gas in kJ $${ mol }^{ -1 }$$ is:

A
15

B
12

C
90

D
8

Solution

Solution:- (A) $$15 \; {kJ}/{mol}$$
As we know that, for ideal gas under any process,
$$\Delta{U} = \cfrac{\text{heat capacity} \times \text{change in temprature}}{\text{no.of moles}}$$
Given:-
Mol. wt. of gas $$= 40 \; g$$
Wt. of gas $$= 5 \; g$$
$$\therefore$$ No. of moles $$= \cfrac{5}{40} = 0.125 \text{ mole}$$
Heat capacity $$= 2.5 \; {kJ}/{K}$$
Change in temperature $$= 298 - 298.75 = 0.75 \; K$$
$$\therefore \Delta{U} = \cfrac{2.5 \times 0.75}{0.125} = 15 \; {kJ}/{mol}$$
Hence the numerical value for the $$\Delta{U}$$ of combustion is $$15 \; {kJ}/{mol}$$.
Question 9
1 / -0

A system is taken along paths A and B as shown. If amounts of heat given in these processes are respectively QA and QB, then:

A
QA=QB

B
QA>QB

C
$$QB<QA$$

D
none of these

Solution

Solution:- (A) $${Q}_{A} = {Q}_{B}$$
As we know that heat is a state function, i.e., it does not depends on the path.
Hence $${Q}_{A} = {Q}_{B}$$
Question 10
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Standard entropy of $${X}_{2},{Y}_{2}$$ and $$X{Y}_{3}$$ are $$60,40$$ and $$50$$ $$J{K}^{-1}$$ $${mol}^{-1}$$, respectively. For the reaction,
$$\cfrac{1}{2}{X}_{2}+\cfrac{3}{2}{Y}_{2}\rightarrow {XY}_{3}.\Delta H=-30kJ$$ to be at equilirbium, the temperature will be:

A
$$1250K$$

B
$$500K$$

C
$$750K$$

D
$$1000K$$

Solution

Solution:- (C) $$750 \; K$$
$$\cfrac{1}{2} {X}_{2} + \cfrac{3}{2} {Y}_{2} \longrightarrow X{Y}_{3} \quad \Delta{H} = -30 \; kJ$$
For the above reaction-
$$\Delta{S} = {\sum{\Delta{S}}}_{\left( Product \right)} - {\sum{\Delta{S}}}_{\left( reactant \right)}$$
$$\Delta{S} = 50 - \left( \cfrac{1}{2} \times 60 + \cfrac{3}{2} \times 40 \right)$$
$$\Rightarrow \Delta{H} = 50 - 90 = -40$$
As we know that,
$$\Delta{G} = \Delta{H} - T \Delta{S}$$
But at equilibrium,
$$\Delta{G} = 0$$
$$\Delta{H} - T \Delta{S} = 0$$
$$\Rightarrow T = \cfrac{\Delta{H}}{\Delta{S}}$$
Given:- $$\Delta{H} = -30 \; kJ = -30000 \; J$$
$$\therefore T = \cfrac{-30000}{-40} = 750 \; K$$
Hence the temperature of the given reaction is $$750 K$$.

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Substrate	$$\Delta H^o$$(kJ/mol)	$$S^o$$(J/mol K)	$$\Delta G^o$$ (kJ/mol)
FeO(s)	$$-266.3$$	$$57.49$$	$$245.12$$
C(Graphite)	$$0$$	$$5.74$$	$$0$$
Fe(s)	$$0$$	$$27.28$$	$$0$$
CO(g)	$$-110.5$$	$$197.6$$	$$-137.15$$

Thermodynamics Test - 66

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Thermodynamics Test - 66

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Question 1

$$\triangle S_{QR} > 0$$

$$\triangle S_{PR} > 0$$

$$\triangle S_{QR} < 0$$

$$\triangle S_{PQ} > 0$$

Solution

Question 2

$$298$$ K

$$668$$ K

$$966$$ K

$$\Delta G^o$$ is $$+$$ve, hence the reaction will never be spontaneous

Solution

Question 3

(i) $$^{-}SH+H_2O\rightarrow H_2S+OH^-$$(ii) $$NH^-_2+H_2O\rightarrow NH_3+OH^-$$

(i) $$^{-}OR+H_2SO_4\rightarrow ROH+HSO^-_4$$(ii) $$R^-+NH_3\rightarrow RH+NH^-_2$$

(i) $$Cl^{-}+HF\rightarrow HCl+F^-$$(ii) $$^-OH+HCl\rightarrow H_2O+Cl^-$$

(i) $$^-OH+HBr\rightarrow H_2O+Br^-$$(ii) $$RO^-+NH_3\rightarrow ROH+NH^-_2$$

Solution

Question 4

$$\Delta H > T \Delta S$$

$$\Delta H < T \Delta S$$

$$\Delta H=T\Delta S$$

$$None\ of\ the\ above$$

Question 5

$$\triangle H>0,\quad \triangle G<0$$

$$\triangle H<0,\quad \triangle S>0$$

$$\triangle H<0,\quad \triangle S<0$$

$$\triangle H>0,\quad \triangle S<0$$

Solution

Question 6

$$\dfrac{1+60\gamma_{Fe}}{1+60\gamma_{Hg}}$$

$$\dfrac{1-60\gamma_{Fe}}{1+60\gamma_{Hg}}$$

$$\dfrac{1+60\gamma_{Fe}}{1-60\gamma_{Hg}}$$

$$\dfrac{1+60\gamma_{Hg}}{1+60\gamma_{Fe}}$$

Solution

Question 7

$$\Delta G=2.30\quad RT\quad log\quad K$$

$$\Delta { G }^{ o }=-2.30\quad RT\quad log\quad K$$

$$\Delta { G }^{ o }=2.30\quad RT\quad log\quad K$$

$$\Delta { G }=-2.30\quad RT\quad log\quad K$$

Solution

Question 8

15

12

90

8

Solution

Question 9

QA=QB

QA>QB

$$QB<QA$$

none of these

Solution

Question 10

$$1250K$$

$$500K$$

$$750K$$

$$1000K$$

Solution

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(i) $$^{-}SH+H_2O\rightarrow H_2S+OH^-$$

(ii) $$NH^-_2+H_2O\rightarrow NH_3+OH^-$$

(i) $$^{-}OR+H_2SO_4\rightarrow ROH+HSO^-_4$$

(ii) $$R^-+NH_3\rightarrow RH+NH^-_2$$

(i) $$Cl^{-}+HF\rightarrow HCl+F^-$$

(ii) $$^-OH+HCl\rightarrow H_2O+Cl^-$$

(i) $$^-OH+HBr\rightarrow H_2O+Br^-$$

(ii) $$RO^-+NH_3\rightarrow ROH+NH^-_2$$