Thermodynamics Test

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Thermodynamics Test - 19

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Question 1
1 / -0

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 :

A
$$-160 kJ$$

B
$$+260 kJ$$

C
$$-2 kJ$$

D
$$+2 kJ$$

Solution

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.
Question 2
1 / -0

The free energy changes for the two reactions given below are:

a. $$SO_2(g) + Cl_2(g)\longrightarrow SO_2Cl_2(g), \Delta G = -2270 cal$$
b. $$S(Rhomb) + O_2(g) + Cl_2(g)\longrightarrow SO_2Cl_2(g), \Delta G = -74060 cal$$

Find $$\Delta G$$ for the reaction $$S(rhom) + O_2(g)\longrightarrow SO_2(g)$$.

A
$$\Delta G = -71790 cal$$

B
$$\Delta G = +71790 cal$$

C
$$\Delta G = -9670 cal$$

D
None of these

Solution

$$S(Rhomb) + O_2(g) + Cl_2(g)\longrightarrow SO_2Cl_2(g), \Delta G = -74060 cal$$....................1
$$SO_2(g) + Cl_2(g)\longrightarrow SO_2Cl_2(g), \Delta G = -2270 cal$$.........................2
so by 2 - 1 we will get
$$\Delta G$$ for the reaction $$S(rhom) + O_2(g)\longrightarrow SO_2(g; \Delta G = -74060 + 2270 = - 71790 cal$$
Question 3
1 / -0

Which law of thermodynamics states that energy can neither be created nor be destroyed, it can change from one to another?

A
First law

B
Second law

C
Zero law

D
None of these

Solution

The Zeroth Law of thermodynamics- If two systems are in thermal equilibrium respectively with a third system, they must be in thermal equilibrium with each other.

First Law-Energy can neither be created nor be destroyed. It can change from one to another. The total energy of the universe remains constant.

Second Law- In a natural thermodynamic process the sum of the entropies of the interacting thermodynamic systems increases.

Third Law-The entropy of a system approaches a constant value as the temperature approaches absolute zero.

Hence option A is the right answer.
Question 4
1 / -0

Hess law is applicable for determination of enthalpy of

A
Reaction

B
Formation

C
Transition

D
All of above

Solution

HESS'S LAW OF CONSTANT HEAT SUMMATION: This law put forth by Hess states that "the resultant enthalpy change in a reaction is the same whether it occurs in one or several steps".
Question 5
1 / -0

Use the bond energies to estimate $$\Delta H$$ for this reaction:

$$H_2(g) + O_2(g)\longrightarrow H_2O_2(g)$$

A
$$-127 kJ$$

B
$$-109 kJ$$

C
$$-400 kJ$$

D
$$-800 kJ$$

Solution

As we know,
$$\Delta H = \sum BE_R - \sum BE_P$$

So, for
$$H_2(g) + O_2(g)\longrightarrow H_2O_2(g)$$\

$$\Delta H = 436+499 - (2\times460 + 142) = -127\ kJ/mol$$
Question 6
1 / -0

A change in the free energy of a system at constant temperature and pressure will be:

$$\Delta_{ sys }G = \Delta_{ sys }H - T\Delta_{ sys }S$$

At constant temperature and pressure
$$\Delta_{ sys }G < 0 (spontaneous)$$
$$\Delta_{ sys }G = 0 (equilibrium)$$
$$\Delta_{ sys }G > 0 (non-spontaneous)$$

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

A
$$-9336 cal$$

B
$$-7006 cal$$

C
$$-2936 cal$$

D
$$+9006 cal$$

Solution

Change in the free energy of a system at constant temperature and pressure will be:

$$\Delta_{ sys }G = \Delta_{ sys }H - T\Delta_{ sys }S$$

As we know,

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

$$\Delta G = 31400- 1273\times32 =- 9336cal $$
Question 7
1 / -0

The heat of hydrogenation of ethene is $$x_1$$ and that of benzene is $$x_2$$
Hence resonance energy of benzene is

A
$$x_1 - x_2$$

B
$$x_1 + x_2$$

C
$$3x_1 - x_2$$

D
$$x_1 - 3x_2$$

Solution

$$CH_2==CH_2 + H_2\longrightarrow CH_3--CH_3\quad\quad \Delta H = x_1$$
$$RE = 3x_1 - x_2$$
Question 8
1 / -0

$$C(diamond) + 2H_2(g)\longrightarrow CH_4(g);\,\,\Delta H_1$$

$$C(g) + 4H(g)\longrightarrow CH_4(g);\,\,\Delta H_2$$

Select the correct relation.

A
$$\Delta H_1 = \Delta H_2$$

B
$$\Delta H_1 > \Delta H_2$$

C
$$\Delta H_1 < \Delta H_2$$

D
$$\Delta H_1 = \Delta H_2 + \Delta_{ vap }H(C) + \displaystyle \Delta_{ diss }H(H_2)$$

Solution

As we know,
$$C(diamond) \longrightarrow C(g)\,\,$$
$$C(diamond) + 2H_2(g)\longrightarrow CH_4(g);\,\,\Delta H_1$$
$$\Delta H_1 = \Delta H_2 + x$$
$$C(g) + 4H(g)\longrightarrow CH_4(g);\,\,\Delta H_2$$
So, $$\Delta H_1 > \Delta H_2 $$
Question 9
1 / -0

$$H_2(g) + Br_2(g)\longrightarrow 2HBr(g); \Delta H^{ \ominus } = -72.40 kJ$$
$$\Delta G^{ \ominus } = -106.49 kJ, T = 298 K$$
The value $$\Delta S$$ is .

A
$$114.3 J k^{ -1 }$$

B
$$-114.3 J k^{ -1 }$$

C
$$1363.6 J k^{ -1 }$$

D
Noe of these

Solution

$$\Delta G = \Delta H - T\Delta S$$
$$ -106.49 kJ = -72.40 - 298\times $$$$\Delta S$$
$$ -34.09 kJ = -298\times$$$$\Delta S$$
$$\Delta S = \frac{ 34090 J }{ 298 } = 114.3 J k^{ -1 }$$
Question 10
1 / -0

A bomb calorimeter is used to measure the value of heat of reaction at a constant

A
Volume

B
Pressure

C
Temperature

D
None of these

Solution

For combustion reactions, we often enclose all reactants in an explosive-proof steel container, called the bomb whose volume does not change during a reaction. The bomb is then submerged in water or other liquid that absorbs the heat of reaction. The heat capacitor of the bomb plus other things is then measured using the same technique as other calorimeters. Such an instrument is called a bomb calorimeter, and its application is called the bomb calorimetry. For combustion reactions, we often enclose all reactants in an explosive-proof steel container, called the bomb whose volume does not change during a reaction. The bomb is then submerged in water or other liquid that absorbs the heat of reaction. The heat capacitor of the bomb plus other things is then measured using the same technique as other calorimeters. Such an instrument is called a bomb calorimeter, and its application is called the bomb calorimetry.
Since volume does not change, a bomb calorimeter measures the heat evolved under constant volume.

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Thermodynamics Test - 19

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Thermodynamics Test - 19

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Question 1

$$-160 kJ$$

$$+260 kJ$$

$$-2 kJ$$

$$+2 kJ$$

Solution

Question 2

$$\Delta G = -71790 cal$$

$$\Delta G = +71790 cal$$

$$\Delta G = -9670 cal$$

None of these

Solution

Question 3

First law

Second law

Zero law

None of these

Solution

Question 4

Reaction

Formation

Transition

All of above

Solution

Question 5

$$-127 kJ$$

$$-109 kJ$$

$$-400 kJ$$

$$-800 kJ$$

Solution

Question 6

$$-9336 cal$$

$$-7006 cal$$

$$-2936 cal$$

$$+9006 cal$$

Solution

Question 7

$$x_1 - x_2$$

$$x_1 + x_2$$

$$3x_1 - x_2$$

$$x_1 - 3x_2$$

Solution

Question 8

$$\Delta H_1 = \Delta H_2$$

$$\Delta H_1 > \Delta H_2$$

$$\Delta H_1 < \Delta H_2$$

$$\Delta H_1 = \Delta H_2 + \Delta_{ vap }H(C) + \displaystyle \Delta_{ diss }H(H_2)$$

Solution

Question 9

$$114.3 J k^{ -1 }$$

$$-114.3 J k^{ -1 }$$

$$1363.6 J k^{ -1 }$$

Noe of these

Solution

Question 10

Volume

Pressure

Temperature

None of these

Solution

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