Thermodynamics Test

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Thermodynamics Test - 65

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Question 1
1 / -0

For the reaction, $$X_{2}O_{4}(l) \rightarrow 2XO_{2}(g)\ \triangle U = 2.1\ k\ cal, \triangle S = 20\ cal K^{-1}$$ at $$300\ K$$. Hence, $$\triangle G$$ is:

A
$$+2.7\ kcal$$

B
$$-2.7\ kcal$$

C
$$+9.3\ kcal$$

D
$$-9.3\ kcal$$

Solution

$${{X}_{2}{O}_{4}}_{\left( l \right)} \longrightarrow 2 {X{O}_{2}}_{\left( g \right)}$$
The change in Gibb's free energy is given by

$$\Delta{G} = \Delta{H} - T \Delta{S}$$
whereas,
$$\Delta{H} =$$ Enthalpy of reaction
$$\Delta{S} =$$ Entropy of reaction $$= 20 \; {cal}/{K}$$

As we know that,
$$\Delta{H} = \Delta{U} + \Delta{{n}_{g}} RT$$
whereas,
$$\Delta{U} =$$ Change in internal energy $$= 2.1 \; kcal = 2100 \; cal$$

$$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = 2 - 0 = 2$$

$$R =$$ Gas constant $$= 2 \; cal$$
$$T = 300 \; K$$

$$\therefore \Delta{H} = 2100 + \left( 2 \times 2 \times 300 \right) = 3300 \; cal$$

$$\therefore \Delta{G} = 3300 - \left( 300 \times 20 \right) = 3300 - 6000 = -2700 cal= -2.7 \; kcal$$
Question 2
1 / -0

For the reaction at $$25, X_{ 2 }O_{ 4 } { O }_{ 4_{ (l) } } \longrightarrow 2X { O }_{ 2_{ (g) } }$$.
$$\Delta H =$$2.1 kcal and $$\Delta S =$$20 cal $${ K }^{ -1 }$$. The reaction would be:

A
spontaneous

B
non-spontaneous

C
at equilibrium

D
unpredictable

Solution

By considering 2 things i.e if reaction is exothermic or endothermic we can justify if reaction is spontaneous or not. If, $$\Delta$$G is negative then reaction is said to be spontaneous. Now, $$\Delta$$G =$$\Delta$$H-$$\Delta$$S. Therefore, $$\Delta$$G will be negative, hence it is a spontaneous reaction.
Question 3
1 / -0

Consider the following processes :-
$$ \delta H (kJ/mol)$$
$$\frac{1}{2} A \rightarrow B + 150$$
$$ 3B + 2C + D - 125$$
$$E + A \rightarrow 2D + 350$$
For $$B + D \rightarrow E + 2C$$ $$\Delta$$$$H$$ will be:

A
$$325 kJ / mol$$

B
$$525 kJ / mol$$

C
$$-175 kJ / mol$$

D
$$-325 kJ / mol$$

Solution
Question 4
1 / -0

Consider the following process
$$\Delta H(kJ/mol)$$
$$\frac{1}{2}A\rightarrow B +50$$
$$3B \rightarrow 3C +D -125$$
$$E + A \rightarrow 2D + 350$$
For $$B + D \rightarrow E +2C, \Delta H $$will be:

A
325 kJ/mol

B
525 kJ/mol

C
-375 kJ/mol

D
-325 kJ/mol

Solution

Solution:-
$$\cfrac{1}{2} A \longrightarrow B + 50$$
$$2 \times \left[ \cfrac{1}{2} A \longrightarrow B + 50 \right]$$
$$A \longrightarrow 2B + 100 ..... \left( 1 \right)$$
$$3B \longrightarrow 2C + D - 125 ..... \left( 2 \right)$$
$$E + A \longrightarrow 2D + 350$$
$$2D \longrightarrow E + A - 350 ..... \left( 3 \right)$$
Adding equation $$\left( 1 \right), \left( 2 \right) \& \left( 3 \right)$$, we have
$$A + 3B + 2D \longrightarrow 2B + 100 + 2C + D - 125 + E + A - 350$$
$$B + D \longrightarrow E + 2C - 375$$
Hence the value of $$\Delta{H}$$ will be $$-375 \; {KJ}/{mol}$$.
Question 5
1 / -0

For the reaction given below the values of standard Gibbs free energy of formation at 298 K are given.
What is the nature of the reaction?
$$I_2 + H_2S \rightarrow 2HI + S$$
$$\Delta G_f^0 (HI) = 1.8 \ kJ\ mol^{-1}$$, $$\Delta G_f^0 (H_2S) = 33.8 \ kJ\ mol^{-1}$$

A
Non-spontaneous in forward direction

B
Spontaneous in forward direction

C
Spontaneous in backward direction

D
Non-spontaneous in both forward and backward directions

Solution

$$I_2 + H_2S \rightarrow 2HI + S$$
$$\Delta G^0 = \sum G_{f(products)}^0 - \sum G_{f(Reactants)}^0 $$ = -30.2 kJ
Hence, the reaction is spontaneous in forward direction.
Question 6
1 / -0

$${ NH }_{ 2 }{ CN }_{ \left( s \right) }+\dfrac { 3 }{ 2 } { O }_{ 2\left( g \right) }\rightarrow { N }_{ 2\left( g \right) }+{ CO }_{ 2\left( g \right) }+{ H }_{ 2 }{ O }_{ \left( l \right) }$$
This reaction is carried out in a bomb calorie-meter. The heat released was $$743\ KJ\ { mol }^{ -1 }$$. The value of $${ \Delta H }_{ 300 }$$ for this reaction would be:

A
$$-740\ KJ\ { mol }^{ -1 }$$

B
$$-741.75\ KJ\ { mol }^{ -1 }$$

C
$$-743.0\ KJ\ { mol }^{ -1 }$$

D
$$-744.25\ KJ\ { mol }^{ -1 }$$

Solution

In a bomb calorimeter, heat released= $$- \Delta U$$
$$\therefore \Delta U= -743 KJmol^{-1}$$
$$\therefore \Delta H= \Delta U-\Delta ngRT$$
where, $$\Delta ng$$= difference between gaseous moles of products and reactants
$$\therefore \Delta ng=(1+1)-\cfrac {3}{2}=\cfrac {1}{2}$$
$$\therefore \Delta H= \Delta U+\cfrac {1}{2} RT$$
$$= -743+\cfrac {1}{2}\times 8.314 \times 300 \times 10^{-3}$$
$$\Delta H_{300}=-741.75 KJ mol^{-1}$$
Question 7
1 / -0

For the combustion of $$CH_4$$ at 1 atm pressure & 300 K, which of the following options is correct?

A
$$\Delta H = \Delta U$$

B
$$\Delta H > \Delta U$$

C
$$ \Delta H < \Delta U$$

D
$$\Delta H = 0$$

Solution

$$\triangle H=\triangle U+\triangle { n }_{ g }RT$$
If $$\triangle { n }_{ g }$$ is positive then $$\triangle H>\triangle U$$, If
$$\triangle { n }_{ g }$$ is negative $$\triangle H<\triangle U$$
Here, $$C{ H }_{ 4(g) }+2{ O }_{ 2\left( g \right) }\rightarrow C{ O }_{ 2\left( g \right) }+2{ H }_{ 2 }{ O }_{ \left( l \right) }$$
$$\Rightarrow \triangle { n }_{ g }=1-\left( 1+2 \right) =-2$$
So, as $$\triangle { n }_{ g }=-ve\Rightarrow \triangle H<\triangle U$$
Question 8
1 / -0

For a given reaction, $$\Delta H = 35.5 kJ mol^{-1}$$ and $$\Delta S = 83.6 kJ mol^{-1}$$. The reaction is spontaneous at: (Assume that $$\Delta H $$ and $$\Delta S$$ do not vary with temperature)

A
T > 425 K

B
all temperatures

C
T > 298 K

D
T < 425 K

Solution

Solution:- (A) $$T > 425 \; K$$
The reaction is $$\begin{cases} \text{spontaneous} & \text{if } \Delta{G} < 0 \\ \text{non-spontaneous} & \text{if } \Delta{G} > 0 \end{cases}$$
As we know that,
$$\Delta{G} = \Delta{H} - T{\Delta{S}}$$
$$\because$$ The reaction is spontaneous.
$$\therefore \Delta{G} < 0$$
$$\Delta{H} - T \Delta{S} < 0$$
$$\Rightarrow T > \cfrac{\Delta{H}}{\Delta{S}}$$
Given:-
$$\Delta{H} = 35.5 \; {KJ}/{mol} = 35500 \; {J}/{mol}$$
$$\Delta{S} = 83.6 \; {J}/{mol-K}$$
$$\therefore T > \cfrac{35500}{83.6}$$
$$\Rightarrow T > 425 \; K$$
Hence the reaction will be spontaneous at $$T > 425 \; K$$.
Question 9
1 / -0

C (diamond) $$\rightarrow$$ C (graphite)
$$\triangle S_{300K} = 10 cal K^{-1}$$
C (diamond) + $$O_{2}$$ $$\rightarrow$$ $$CO_{2}$$
$$\triangle$$ H = -91 cal $$mol^{-1}$$ at 300 K
C (graphite) + $$O_{2}$$ $$\rightarrow$$ $$CO_{2}$$
$$\triangle$$ H = X at 300 K
X is:

A
-81 K cal

B
101 K cal

C
-94 K cal

D
88 K cal

Solution
Question 10
1 / -0

Which thermochemical law is represented by the following figure?

A
Standard enthalpy of a reaction

B
Born - Haber cycle of lattice enthalpy

C
Hess's law of constant heat summation

D
Standard enthalpy of a solution

Solution

According to Hess' Law of Constant Heat Summation,
If $$ X \xrightarrow {\Delta H} Y$$,
And $$X \xrightarrow {\Delta H_1} P \xrightarrow {\Delta H_2} Q \xrightarrow {\Delta H_3} Y$$
then ,$$ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3$$

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Thermodynamics Test - 65

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Thermodynamics Test - 65

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Question 1

$$+2.7\ kcal$$

$$-2.7\ kcal$$

$$+9.3\ kcal$$

$$-9.3\ kcal$$

Solution

Question 2

spontaneous

non-spontaneous

at equilibrium

unpredictable

Solution

Question 3

$$325 kJ / mol$$

$$525 kJ / mol$$

$$-175 kJ / mol$$

$$-325 kJ / mol$$

Solution

Question 4

325 kJ/mol

525 kJ/mol

-375 kJ/mol

-325 kJ/mol

Solution

Question 5

Non-spontaneous in forward direction

Spontaneous in forward direction

Spontaneous in backward direction

Non-spontaneous in both forward and backward directions

Solution

Question 6

$$-740\ KJ\ { mol }^{ -1 }$$

$$-741.75\ KJ\ { mol }^{ -1 }$$

$$-743.0\ KJ\ { mol }^{ -1 }$$

$$-744.25\ KJ\ { mol }^{ -1 }$$

Solution

Question 7

$$\Delta H = \Delta U$$

$$\Delta H > \Delta U$$

$$ \Delta H < \Delta U$$

$$\Delta H = 0$$

Solution

Question 8

T > 425 K

all temperatures

T > 298 K

T < 425 K

Solution

Question 9

-81 K cal

101 K cal

-94 K cal

88 K cal

Solution

Question 10

Standard enthalpy of a reaction

Born - Haber cycle of lattice enthalpy

Hess's law of constant heat summation

Standard enthalpy of a solution

Solution

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