Thermodynamics Test

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Thermodynamics Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

Reaction of silica with mineral acids may be given as:
$$Si{ O }_{ 2 }+4HF\longrightarrow Si{ F }_{ 4 }+2{ H }_{ 2 }O;\Delta H=-10.17kcal$$
$$Si{ O }_{ 2 }+4HCl\longrightarrow Si{ Cl }_{ 4 }+2{ H }_{ 2 }O;\Delta H=+36.7kcalHCl$$
Which among the following is correct?

A
$$HF$$ and $$HCl$$ both will react with silica

B
Only $$HF$$ will react with silica

C
Only $$HCl$$ will react with silica

D
Neither $$HF$$ nor $$HCl$$ will react with silica

Solution

When silica reacts with $$HCl$$ then the reaction is endothermic and heat is absorbed by the system, when silica reacts with $$HF$$ we get $$\Delta H = -ve$$, i.e., exothermic reaction and hence the reaction become feasible,
so, we can say that only $$HF$$ reacts with silica.
Question 2
1 / -0

Which one of the following ores is best concentrated by froth floatation method?

A
Magnetite

B
Galena

C
Malechitee

D
none

Solution

Froth flotation method is used for the concentration of sulphide ores. The method is based on the preferential wetting properties with the frothing agent and water. In the given options, Galena (PbS) being a sulphide ore is concentrated by froth floatation process.
Question 3
1 / -0

Predict the sign of $$\Delta S$$ for each of the following processes, which occur at constant temperature:
$$I$$. The volume of $$2$$ mol of $${O}_{2}(g)$$ increases from $$44L$$ to $$54L$$
$$II$$. The pressure of $$2$$ mil of $${O}_{2}(g)$$ increases from $$1$$ atm to $$1.2$$ atm.

A
$$(I)\rightarrow \Delta S=-ve$$; $$(II)\rightarrow \Delta S=-ve$$

B
$$(I)\rightarrow \Delta S=-ve$$; $$(II)\rightarrow \Delta S=+ve$$

C
$$(I)\rightarrow \Delta S=+ve$$; $$(II)\rightarrow \Delta S=-ve$$

D
$$(I)\rightarrow \Delta S=+ve$$; $$(II)\rightarrow \Delta S=+ve$$

Solution

I. On increasing the volume from $$44L\, to\, 54L$$ the degree of randomization increases and hence, entropy also increases,
so we get $$\Delta S = +ve$$
II. The entropy of the system decreases on increasing the pressure, so we get
$$\Delta S = -ve$$
Question 4
1 / -0

From the following data of heats of combustion, find the heat of formation of $${ CH }_{ 3 }OH(l)$$:
$${ CH }_{ 3 }OH(l)+\cfrac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow C{ O }_{ 2 }(g)+2{ H }_{ 2 }O(l);\Delta H=-726kJ$$
$$C(s)+{ O }_{ 2 }(g)\longrightarrow C{ O }_{ 2 }(g);\Delta H=-394kJ$$
$${ H }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow { H }_{ 2 }O(l);\Delta H=-286kJ$$

A
$$240\ kJ\ { mol }^{ -1 }$$

B
$$-240\ kJ\ { mol }^{ -1 }$$

C
$$-140\ kJ\ { mol }^{ -1 }$$

D
$$140\ kJ\ { mol }^{ -1 }$$

Solution

Heat of formation of $$CH_3OH$$ = 2$$\times $$ heat of combustion of $$H_2O$$ + heat of combustion of $$CO_2$$ - heat of combustion of $$CH_3OH$$
Heat of formation of $$CH_3OH$$ = $$2*(-286) - (-394) - (-726) = -240 KJmol^{-1}$$
Question 5
1 / -0

Directions For Questions

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The free energy for a reaction having $$\Delta H=31400\ cal,\Delta S=32\ cal\ { K }^{ -1 }{mol}^{-1}$$ at $${1000}^{o}C$$ is:

A
$$-9336\ cal$$

B
$$-7386\ cal$$

C
$$-1936\ cal$$

D
$$+9336\ cal$$

Solution

We have,
$$\Delta G = \Delta H -T\Delta S$$
$$\Delta G = 31400 - 1273*32 = -9336 cal$$
Question 6
1 / -0

The most abundant element in the universe is thought to be:

A
Hydrogen

B
Carbon

C
Oxygen

D
Nitrogen

Solution

Hydrogen is the most abundant element in the universe followed by Helium.
Question 7
1 / -0

Calculate the heat of formation of methane, given that
heat of formation of water $$=-286kJ\quad { mol }^{ -1 }$$
heat of combustion of methane $$=-890kJ\quad { mol }^{ -1 }$$
heat of combustion of carbon $$=-393.5kJ\quad { mol }^{ -1 }$$

A
$$75.5\ kJ\ { mol }^{ -1 }$$

B
$$-75.5\ kJ\ { mol }^{ -1 }$$

C
$$-55.5\ kJ\ { mol }^{ -1 }$$

D
$$55.5\ kJ\ { mol }^{ -1 }$$

Solution

Heat of formation of methane = 2$$\times$$ heat of formation of water + heat of combustion of carbon - heat of combustion of methane
Heat of formation of methane = $$2*(-286) + (-393.5) -(-890)$$ =$$-75.5 KJmol^{-1}$$
Question 8
1 / -0

When a substance A reacts with water it produces a combustible gas B and a solution of substance C in water. When another substance D reacts with this solution of C, it also produces the same gas B on warming but D can produce gas B on reaction with dilute sulphuric acid at room temperature. A imparts a deep golden yellow colour to a smokeless flame of Bunsen burner. A, B, C and D respectively are:

A
$$Na,H_2,NaOH,Zn$$

B
$$K,H_2,KOH,Al$$

C
$$Ca,H_2,Ca(OH)_2,Sn$$

D
$$None$$

Solution

solution:
$$2Na + 2H_2O \rightarrow H2 +2ZnO_2 + 2NaOH$$
A B C
and $$Zn + NaOH \rightarrow Na_2ZnO_2 + H_2Zn+H2SO4(dil.) \rightarrow ZnSO_4 + H2$$
D C
only from this part we can say that correct answer is A
Question 9
1 / -0

Calculate the standard free energy change for the combustion of glucose at $$298\ K$$, using the given data.
$${ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }+6{ O }_{ 2 }(g)\longrightarrow 6{ CO }_{ 2 }(g)+6{H}_{2}O$$
$$\Delta { H }^{ o }=-2820\ kJ\ { mol }^{ -1 };\Delta { S }^{ o }=210\ J{ K }^{ -1 }{ mol }^{ -1 }$$

A
$$-1882.58\ kJ\ { mol }^{ -1 }$$

B
$$-2882.58\ kJ\ { mol }^{ -1 }$$

C
$$-3882.58\ kJ\ { mol }^{ -1 }$$

D
$$-4882.58\ kJ\ { mol }^{ -1 }$$

Solution

For standard free energy change we have
$$\Delta G = \Delta H - T* \Delta S$$
$$\Delta G = -2820000 - 298* 210 = -2882580 Jmol^{-1}$$
$$\Delta G=-2882.58 KJmol^{-1}$$
Question 10
1 / -0

For the reaction,
$${ Br }_{ 2 }(l)+{ Cl }_{ 2 }(g)\longrightarrow 2BrCl(g)$$
$$\Delta H=29.37kJ\quad { mol }^{ -1 };\Delta S=104J\quad { mol }^{ -1 }$$. Find the temperature above which the reaction would become spontaneous.

A
Above $$273.4\ K$$

B
Above $$283.4\ K$$

C
Above $$282.4\ K$$

D
Above $$373.4\ K$$

Solution

We have,
$$\Delta G = \Delta H - T\Delta S$$
so, spontaneous reaction we have $$\Delta G <0$$
so $$ \Delta H - T\Delta S <0$$
$$ 29370 - T*104 <0$$
so, temperature should be above 282.4 K

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Thermodynamics Test - 31

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Thermodynamics Test - 31

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Question 1

$$HF$$ and $$HCl$$ both will react with silica

Only $$HF$$ will react with silica

Only $$HCl$$ will react with silica

Neither $$HF$$ nor $$HCl$$ will react with silica

Solution

Question 2

Magnetite

Galena

Malechitee

none

Solution

Question 3

$$(I)\rightarrow \Delta S=-ve$$; $$(II)\rightarrow \Delta S=-ve$$

$$(I)\rightarrow \Delta S=-ve$$; $$(II)\rightarrow \Delta S=+ve$$

$$(I)\rightarrow \Delta S=+ve$$; $$(II)\rightarrow \Delta S=-ve$$

$$(I)\rightarrow \Delta S=+ve$$; $$(II)\rightarrow \Delta S=+ve$$

Solution

Question 4

$$240\ kJ\ { mol }^{ -1 }$$

$$-240\ kJ\ { mol }^{ -1 }$$

$$-140\ kJ\ { mol }^{ -1 }$$

$$140\ kJ\ { mol }^{ -1 }$$

Solution

Question 5

$$-9336\ cal$$

$$-7386\ cal$$

$$-1936\ cal$$

$$+9336\ cal$$

Solution

Question 6

Hydrogen

Carbon

Oxygen

Nitrogen

Solution

Question 7

$$75.5\ kJ\ { mol }^{ -1 }$$

$$-75.5\ kJ\ { mol }^{ -1 }$$

$$-55.5\ kJ\ { mol }^{ -1 }$$

$$55.5\ kJ\ { mol }^{ -1 }$$

Solution

Question 8

$$Na,H_2,NaOH,Zn$$

$$K,H_2,KOH,Al$$

$$Ca,H_2,Ca(OH)_2,Sn$$

$$None$$

Solution

Question 9

$$-1882.58\ kJ\ { mol }^{ -1 }$$

$$-2882.58\ kJ\ { mol }^{ -1 }$$

$$-3882.58\ kJ\ { mol }^{ -1 }$$

$$-4882.58\ kJ\ { mol }^{ -1 }$$

Solution

Question 10

Above $$273.4\ K$$

Above $$283.4\ K$$

Above $$282.4\ K$$

Above $$373.4\ K$$

Solution

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