Thermodynamics Test

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Thermodynamics Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

$$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -286 kJ$$
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l)......... kJ(\pm?)$$

A
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 572 kJ$$.

B
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) - 572 kJ$$.

C
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) +0 kJ$$.

D
None of these

Solution

Reaction:
$$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -286 kJ$$.................1
multiply it by 2 we will get,
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l); \Delta H = -286*2 = -572 kJ$$
which can be written as
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 572 kJ$$.
Question 2
1 / -0

The heat released when the requisite amount of ions in the gaseous state combine to give 1 mol of crystal lattice is known as:

A
lattice energy

B
hydration energy

C
formation energy

D
none of the above

Solution

The lattice energy of a crystalline solid is usually defined as the energy of formation of the crystal from infinitely-separated ions, molecules, or atoms.
The heat energy released when new bonds are made between the ions and water molecules is known as the hydration enthalpy.
The standard enthalpy of formation of a compound is the change of enthalpy from the formation of 1 mole of the compound from its constituent elements.
Hence, option A is the right answer.
Question 3
1 / -0

Directions For Questions

The industrial preparation of a polymer, PTFE, is based on the synthesis of the monomer $${ CF }_{ 2 }={ CF }_{ 2 }$$, which is produced according to reaction (i) below
$${ 2CHClF }_{ 2 }$$(g) $$\longrightarrow $$ $${ CF }_{ 2 }$$ = $${ CF }_{ 2 }(g)$$ $$+ 2HCl$$
The monomer $${ CF }_{ 2 }$$ = $${ CF }_{ 2 }$$ is also obtained by reaction (ii) below
$${ 2CF }_{ 3 }$$(g) $$\longrightarrow $$ $${ CF }_{ 2 }$$ = $${ CF }_{ 2 }(g)$$ + $$2HF(g)$$ ; $$\Delta $$H = 198.1 kJ/mol

...view full instructions

The enthalpy change for reaction (i) is:

A
+ 100.23 kJ/mol

B
+ 127.5 kJ/mol

C
- 127.5 kJ/mol

D
- 100.23 kJ/mol

Solution

$${ \Delta H }_{ r }$$ =$$[({ \Delta H }_{ f })_{ C_{ 2 }{ F }_{ 4 } }]+2({ \Delta H }_{ f })_{ HCl }-2({ \Delta H }_{ f })_{ ^{ { CHClF }_{ 2 } } }]$$
$$= [658.3 + 2(92.3) + 2(485.2)]$$
$$= 127.5\ kJ/mole$$
Question 4
1 / -0

Gibbs- Helmholtz equation is

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

B
$$\Delta G = \Delta H +T\Delta S$$.

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

D
none of these

Solution

Gibbs-Helmholtz Equation
$$\Delta G=\Delta H-T\Delta S$$
Measure of heat in the reaction
$$\bullet $$ In chemical reactions, reactions absorb energy to break bonds
$$\bullet $$ Energy is then released when bonds form between rearranged atoms of the product
Question 5
1 / -0

The relationship between the free energy change $$(\Delta G)$$ and entropy change $$(\Delta S)$$ at constant temperature $$(T)$$ is

A
$$\,\Delta G = \Delta H + T\Delta S$$

B
$$\,\Delta H = \Delta G + T\Delta S$$

C
$$\,T\Delta S = \Delta G + \Delta H$$

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

Solution

As we know,
$$\,\Delta G = \Delta H - T\Delta S$$
So,
$$\,\Delta H = \Delta G + T\Delta S$$
Question 6
1 / -0

What can be concluded about the values of $$\Delta H$$ and $$\Delta S$$ from this graph ?

A
$$\Delta H >0, \:\Delta S >0$$

B
$$\Delta H >0, \:\Delta S <0$$

C
$$\Delta H <0, \:\Delta S >0$$

D
$$\Delta H <0, \:\Delta S <0$$

Solution

As we know,
$$\Delta G = \Delta H - T \Delta S$$
here,
as slope is negative so $$- T \Delta S < 0$$
&
$$ T \Delta S > 0$$
$$\Delta S >0$$
Also as intercept is positive so
$$ \Delta H > 0$$
Question 7
1 / -0

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v : $${ \Delta H }_{ atomisation }$$ K = 90 kJ/mol
w : $${ \Delta H }_{ ionisation }$$ K = 418 kJ/mol
x : $${ \Delta H }_{ dissociation }$$ H = 436 kJ/mol
y : $${ \Delta H }_{ electron affinity }$$ H = 78 kJ/mol
z : $${ \Delta H }_{ lattice }$$ KH = 710 kJ/mol
In terms of the letters v to z the expression for
$${ \Delta H }_{ i }$$ of K is $${ \Delta H }_{ i }$$ = $$w/2$$.
If true enter 1, else enter 0.

A
0

B
1

C
2

D
3
Question 8
1 / -0

The Born Haber cycle below represents the energy changes occurring at 298K when $$KH$$ is formed from its elements
v : $${ \Delta H }_{ atomisation }$$ $$K = 90 kJ/mol$$
w : $${ \Delta H }_{ ionisation }$$ $$K = 418 kJ/mol$$
x : $${ \Delta H }_{ dissociation }$$ $$H = 436 kJ/mol$$
y : $${ \Delta H }_{ electron affinity }$$ $$H = 78 kJ/mol$$
z : $${ \Delta H }_{ lattice }$$ $$KH = 710 kJ/mol$$
Calculate the value of $$\Delta $$$$H$$ showing all your working.

A
124 kJ/mol

B
-124 kJ/mol

C
124 J/mol

D
None of these

Solution

Born-Haber Cycle
It is a series of steps (chemical processes) used to calculate the lattice energy of ionic solids, which is difficult to determine experimentally. You can think of BH cycle as a special case of Hess's law which states that the overall energy change in a chemical process can be calculated by breaking down the process into several steps and adding the energy change from each step.
$${ \Delta H }_{ r }$$=2 $$\times $$ 90 + 2 $$\times $$ 418 + 436 - 2 $$\times $$ 78 - 2 $$\times $$ 710
$${ \Delta H }_{ r }$$= $$- 124 kJ/mole$$
Question 9
1 / -0

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v : $${ \Delta H }_{ atomisation }$$ K = 90 kJ/mol
w : $${ \Delta H }_{ ionisation }$$ K = 418 kJ/mol
x : $${ \Delta H }_{ dissociation }$$ H = 436 kJ/mol
y : $${ \Delta H }_{ electron affinity }$$ H = 78 kJ/mol
z : $${ \Delta H }_{ lattice }$$ KH = 710 kJ/mol
In terms of the letters v to z the expression for
$${ \Delta H }_{ electron affinity }$$ of H is $${ \Delta H }_{ electron affinity }$$ is __.
I

A
$$y$$

B
$$y/2$$

C
$$2y$$

D
$$y/3$$

Solution

The electron affinity $$\Delta { H }_{ electron\: affinity }$$ of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion. It is exothermic as heat energy is released from the system.

$$\Delta { H }_{ electron\: affinity }$$ should be y not y/2.
Question 10
1 / -0

What is the free energy change $$(\Delta G)$$ when 1.0 mole of water at $$100^{\circ}\! C$$ and 1 atm pressure is converted into steam at $$100^{\circ}\! C$$ and 1 atm pressure?

A
80 cal

B
540 cal

C
620 cal

D
zero

Solution

$$H_2O_{(l)} \longrightarrow H_2O_{(g)}$$
373 K 373 K
1atm 1atm

$$\displaystyle \Delta S=\frac{\Delta H_{vap}}{T}$$

$$\displaystyle \Delta G=\Delta H_f - \Delta H_i=0$$

Hence, the correct option is $$D$$

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Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 20

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Thermodynamics Test - 20

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SHARING IS CARING

Question 1

$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 572 kJ$$.

$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) - 572 kJ$$.

$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) +0 kJ$$.

None of these

Solution

Question 2

lattice energy

hydration energy

formation energy

none of the above

Solution

Question 3

+ 100.23 kJ/mol

+ 127.5 kJ/mol

- 127.5 kJ/mol

- 100.23 kJ/mol

Solution

Question 4

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

$$\Delta G = \Delta H +T\Delta S$$.

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

none of these

Solution

Question 5

$$\,\Delta G = \Delta H + T\Delta S$$

$$\,\Delta H = \Delta G + T\Delta S$$

$$\,T\Delta S = \Delta G + \Delta H$$

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

Solution

Question 6

$$\Delta H >0, \:\Delta S >0$$

$$\Delta H >0, \:\Delta S <0$$

$$\Delta H <0, \:\Delta S >0$$

$$\Delta H <0, \:\Delta S <0$$

Solution

Question 7

0

1

2

3

Question 8

124 kJ/mol

-124 kJ/mol

124 J/mol

None of these

Solution

Born-Haber Cycle

Question 9

$$y$$

$$y/2$$

$$2y$$

$$y/3$$

Solution

Question 10

80 cal

540 cal

620 cal

zero

Solution

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