Thermodynamics Test

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Thermodynamics Test - 39

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Question 1
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An imaginary reaction $$X\rightarrow Y$$ takes place in three steps $$X\rightarrow A,\Delta H=-q_{1}$$; $$B\rightarrow A,\Delta H=-q_{2}$$; $$Y\rightarrow B,\Delta H=-q_{3}$$ If Hess' law is applicable, then the heat of the reaction $$ X \rightarrow Y $$ is:

A
$$ q_{1}-q_{2}+q_{3} $$

B
$$ q_{2}-q_{1}+q_{3} $$

C
$$ q_{1}-q_{2}-q_{3} $$

D
$$ q_{3}-q_{2}-q_{1} $$

Solution

Solution:-
$$X \longrightarrow A \quad \Delta{H} = - {q}_{1} ..... \left( 1 \right)$$
$$B \longrightarrow A \quad \Delta{H} = - {q}_{2}$$
$$A \longrightarrow B \quad \Delta{H} = {q}_{2} ..... \left( 2 \right)$$
$$Y \longrightarrow B \quad \Delta{H} = - {q}_{3}$$
$$B \longrightarrow Y \quad \Delta{H} = {q}_{3} ..... \left( 3 \right)$$
Hess's Law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
Therefore,
Adding $${eq}^{n} \left( 1 \right), \left( 2 \right) \& \left( 3 \right)$$, we have
$$X + A + B \longrightarrow A + B + Y \quad \Delta{H} = \left( -{q}_{1} \right) + \left( {q}_{2} \right) + \left( {q}_{3} \right)$$
$$X \longrightarrow Y \quad \Delta{H} = {q}_{2} - {q}_{1} + {q}_{3}$$
Hence the heat of the reaction $$X \rightarrow Y$$ is $${q}_{2} - {q}_{1} + {q}_{3}$$.
Question 2
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The spontaneous nature of a reaction is impossible only if:

A
$$\Delta H=+{ve}, \Delta S=+{ve}$$

B
$$\Delta H=-{ve}, \Delta S=-{ve}$$

C
$$\Delta H=-{ve}, \Delta S=+{ve}$$

D
$$\Delta H=+{ve}, \Delta S=-{ve}$$

Solution

Using the relation, $$\triangle G= \triangle H - T \triangle S$$

$$\triangle G$$ is positive at all conditions of T and P only when $$\triangle H$$ is positive and $$\triangle S$$ is negative.

Since $$\triangle G$$ is positive. The reaction is non-spontaneous which is not possible.

So, the correct option is $$D$$
Question 3
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Which of the following set of condition necessarily implies a spontaneous process?

A
Exothermic reaction with increasing disorder

B
Endothermic reaction with increasing disorder

C
Exothermic reaction with decreasing disorder

D
Endothermic reaction with decreasing disorder

Solution

Solution:- (A) Exothermic reaction with increasing disorder
For a spontaneous reaction
$$\Delta{H} = -ve$$
$$\Delta{S} = +ve$$
Spontaneity depends upon both critical minimum energy and maximum randomness or disorder.
Question 4
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Which of the natural process is spontaneous?

A
Formation of curd from milk after doing initiation

B
Convesion of $$C (graphite) \rightarrow C (diamond)$$ at $$25^0$$C and $$1$$atm

C
Formation of $$H_2$$ (g) and $$O_2$$(g) from $$H_2O$$(l)

D
Formation of $$CrO_5$$ from $$H_2O_2$$ and $$K_2Cr_2O_7$$ in basic medium

Solution

(A) The curd forms from milk because of the chemical reaction between the lactic acid, bacteria and casein. During fermentation, the bacteria uses enzymes to produce energy $$(ATP)$$ from lactose. This is spontaneous.
(B) $$C(graphite)\rightarrow C(diamond)$$ at $$25^oC$$ and $$1 atm$$. It is not spontaneous. It needs a $$150000$$ times positive atmospheric pressure at Earth's surface to do so.
(C) It is not spontaneous. Only electric discharge can convert $${H_2}_{(g)}$$ and $${O_2}_{(g)}$$ to $${H_2O}_{(l)}$$ .
(D) It is not sponatneous. Acidified solution of $$K_2Cr_2O_7$$ reacts with $$H_2O_2$$ to form $$CrO_5$.
$$Cr_2O_7^{2-}+2H^++H_2O_2 \longrightarrow CrO_5+5H_2O$$
Question 5
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Choose the correct statement.

A
All spontaneous natural processes are thermodynamically irreversible.

B
Work done is a state function.

C
Work done is zero during an adiabatic expansion.

D
For a spontaneous process it is always true that $$\Delta S_{system}>0$$.
Question 6
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Consider the following reaction,
$$2A + B \rightarrow C + 2D$$, $$\Delta H_{1} = 10$$
$$A + 2C \rightarrow 2D + B$$, $$\Delta H_{2} = -5$$ What is $$\Delta H$$ of reaction $$A + 2B \rightarrow 3C$$?

A
$$-5$$

B
$$+5$$

C
$$+10$$

D
$$+15$$

Solution

Solution:- (D) $$+15$$
$$A + 2B \longrightarrow 3C \quad \Delta{H} = ?$$
Given:-
$$2A + B \longrightarrow C + 2 D \quad \Delta{{H}_{1}} = 10 ..... \left( 1 \right)$$
$$A + 2C \longrightarrow 2D + B \quad \Delta{{H}_{2}} = -5$$
$$B + 2D \longrightarrow A + 2C \quad \Delta{{H}_{3}} = - \Delta{{H}_{2}} = 5 ..... \left( 2 \right)$$
Adding $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have
$$2A + B + B + 2D \longrightarrow C + 2D + A + 2C \quad \Delta{H} = 10 + 5$$
$$A + 2B \longrightarrow 3C \quad \Delta{H} = 15$$
Hence the $$\Delta{H}$$ for the given reaction is $$+15$$.
Question 7
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At $$300 \,K$$, the reactions which have following values of thermodynamic occurs spontaneously

A
$$\Delta G^o = -400 \,kJ \,mol^{-1}$$

B
$$\Delta H^o = -200 \,kJ \,mol^{-1}, \Delta S^o = 4 \,kJK^{-1}mol^{-1}$$

C
$$\Delta H^o = 200 \,kJ \,mol^{-1}, \Delta S^o = 4k \,JK^{-1} mol^{-1}$$

D
all of these

Solution

For spontaneous reaction,

Gibbs free energy, $$\Delta{G}<0$$

$$\Delta S = +ve$$ $$\Delta G = -ve$$

$$\Delta H = -ve$$

(i) $$\Delta G = \Delta H - T \Delta S$$

$$\Delta{G}<O$$

(ii) $$\Delta G = -200 - 300 \times 4=-1400 $$

$$\Delta G = -ve$$

(iii) $$\Delta G = 200 - 300 \times 4$$

$$\Delta G = -1000 - ve$$

All the given options suggest a spontaneous reaction.

Hence, the correct option is D.
Question 8
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Consider the following sequence of reactions
Which of the following is always true?

A
$$\Delta H_1=\Delta H_2+\Delta H_3+\Delta H_4$$

B
$$\Delta H_2=\Delta H_1+\Delta H_3+\Delta H_4$$

C
$$\Delta H_3=\Delta H_1+\Delta H_2+\Delta H_4$$

D
$$\Delta H_4=\Delta H_1+\Delta H_2+\Delta H_3$$

Solution

Solution:- (B) $$\Delta{{H}_{2}} = \Delta{{H}_{1}} + \Delta{{H}_{3}} + \Delta{{H}_{4}}$$
According to Hess's Law, regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
Hence,
$$\Delta{{H}_{2}} = \Delta{{H}_{1}} + \Delta{{H}_{3}} + \Delta{{H}_{4}}$$
Question 9
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The pressure at the base of a column of liquid of length / and held at an angle $$\theta $$ to the vertical is

A
$$\rho gl$$

B
$$\rho gl \sin { \theta }$$

C
$$\rho g\cos { \theta } $$

D
$$\rho gl\cos { \theta } $$

Solution

Solutiion:- (D) $$\rho g l \cos{\theta}$$
The pressure at the base of a column of liquid of length $$l$$ and held at an angle $$\theta$$ to the vertical is given as-
$$P = \rho g l \cos{\theta}$$
Question 10
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Calculate the amount of heat necessary to raise $$213.5 g$$ of water from $${25}^{o}$$ and $${100}^{o}C$$ Molar heat capacity of water is $$18 cal$$ $${mole}^{-1}{K}^{-1}$$

A
$$16.0125 kcal$$

B
$$12.0125 kcal$$

C
$$20.222 kcal$$

D
$$26.323 kcal$$

Solution

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Thermodynamics Test - 39

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Thermodynamics Test - 39

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Question 1

$$ q_{1}-q_{2}+q_{3} $$

$$ q_{2}-q_{1}+q_{3} $$

$$ q_{1}-q_{2}-q_{3} $$

$$ q_{3}-q_{2}-q_{1} $$

Solution

Question 2

$$\Delta H=+{ve}, \Delta S=+{ve}$$

$$\Delta H=-{ve}, \Delta S=-{ve}$$

$$\Delta H=-{ve}, \Delta S=+{ve}$$

$$\Delta H=+{ve}, \Delta S=-{ve}$$

Solution

Question 3

Exothermic reaction with increasing disorder

Endothermic reaction with increasing disorder

Exothermic reaction with decreasing disorder

Endothermic reaction with decreasing disorder

Solution

Question 4

Formation of curd from milk after doing initiation

Convesion of $$C (graphite) \rightarrow C (diamond)$$ at $$25^0$$C and $$1$$atm

Formation of $$H_2$$ (g) and $$O_2$$(g) from $$H_2O$$(l)

Formation of $$CrO_5$$ from $$H_2O_2$$ and $$K_2Cr_2O_7$$ in basic medium

Solution

Question 5

All spontaneous natural processes are thermodynamically irreversible.

Work done is a state function.

Work done is zero during an adiabatic expansion.

For a spontaneous process it is always true that $$\Delta S_{system}>0$$.

Question 6

$$-5$$

$$+5$$

$$+10$$

$$+15$$

Solution

Question 7

$$\Delta G^o = -400 \,kJ \,mol^{-1}$$

$$\Delta H^o = -200 \,kJ \,mol^{-1}, \Delta S^o = 4 \,kJK^{-1}mol^{-1}$$

$$\Delta H^o = 200 \,kJ \,mol^{-1}, \Delta S^o = 4k \,JK^{-1} mol^{-1}$$

all of these

Solution

Question 8

$$\Delta H_1=\Delta H_2+\Delta H_3+\Delta H_4$$

$$\Delta H_2=\Delta H_1+\Delta H_3+\Delta H_4$$

$$\Delta H_3=\Delta H_1+\Delta H_2+\Delta H_4$$

$$\Delta H_4=\Delta H_1+\Delta H_2+\Delta H_3$$

Solution

Question 9

$$\rho gl$$

$$\rho gl \sin { \theta }$$

$$\rho g\cos { \theta } $$

$$\rho gl\cos { \theta } $$

Solution

Question 10

$$16.0125 kcal$$

$$12.0125 kcal$$

$$20.222 kcal$$

$$26.323 kcal$$

Solution

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