Thermodynamics Test

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Thermodynamics Test - 34

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Question 1
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For the reaction : $${ Cl }_{ 2 }(g) \longrightarrow 2{ Cl }(g),$$ _________

A
$$\Delta$$ H is positive, $$\Delta$$ S is positive

B
$$\Delta$$ H is positive, $$\Delta$$ S is negative

C
$$\Delta$$ H is negative, $$\Delta$$ S is negative

D
$$\Delta$$ H is negative, $$\Delta$$ S is positive

Solution

$$\Delta$$H and $$\Delta$$S are negative.
The given reaction represents the formation of chlorine molecule from chlorine atoms.
Here, the bond formation is taking place. Therefore, energy is being released. Hence, ∆H is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ∆S is negative for the given reaction.
Hence option C is the correct answer.
Question 2
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$$\Delta$$G is the net energy available to do useful work and is a measure of free energy. If a reaction has positive enthalpy change and positive entropy change, under what conditions will the reaction be spontaneous?

A
$$\Delta$$G will be positive at low temperature hence reaction is spontaneous at low temperature.

B
$$\Delta$$G will be negative at high temperature hence reaction is spontaneous at high temperature.

C
$$\Delta$$G will be negative at low temperature hence reaction is spontaneous at low temperature.

D
$$\Delta$$G will be negative at all temperature hence reaction is spontaneous at all temperatures.

Solution

$$\Delta H>0$$ and $$\Delta S>0$$
Now, $$\Delta G= \Delta H- T \Delta S$$
For a spontaneous process, $$\Delta G<0,i.e, T\Delta S > \Delta H$$, i.e., temperature should be high.
Therefore, the reaction will be spontaneous for negative value of $$\Delta G$$, i.e., high temperature is required.
Question 3
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Hess's law is applicable for the determination of heat of ________.

A
transition

B
formation

C
reaction

D
all of these.

Solution

Hess's Law of Constant Heat Summation: The standard enthalpy of a reaction, which takes place in several steps, is the sum of the standard enthalpies of intermediate reactions into which the overall reactions may be divided at the same temperature.
$$\Delta H=\Delta { H }_{ 1 }+\Delta { H }_{ 2 }+\Delta { H }_{ 3 }$$
It is applicable to determine heat of formation, transition, reaction, hydration, etc.
Question 4
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Which of the following statements regarding Gibb's energy change is correct?

A
If $$\Delta$$G is negative (< 0), the process is non-spontaneous.

B
If $$\Delta$$G is positive (> 0), the process is non-spontaneous.

C
If $$\Delta$$G is negative (< 0), the process is spontaneous.

D
If $$\Delta$$G is positive (> 0), the process is equilibrium.

Solution

The energy available for a system at some conditions and by which useful work can be done is Gibb's energy (G).
$$G=H-TS$$
For spontaneous process, $$\Delta G<0$$
For non-spontaneous process, $$\Delta G>0$$
For equilibrium process, $$\Delta G=0$$
Question 5
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The total entropy change ($$\Delta S_{total}$$) for the system and surroundings of a spontaneous process is given by :

A
$$\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} > 0$$

B
$$\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} < 0$$

C
$$\Delta S_{system} = \Delta S_{total} + \Delta S_{surr} > 0$$

D
$$\Delta S_{surr} = \Delta S_{total} + \Delta S_{system} < 0$$

Solution

According to Second Law Of Thermodynamics, for a spontaneous process, total change in entropy (system + surrounding) should be positive.
Question 6
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$$\Delta$$ U = q + w, is mathematical expression for:

A
first law of thermodynamics

B
second law of thermodynamics

C
third law of thermodynamics

D
zeroth law of thermodynamics

Solution

First law of thermodynamics is represented mathematically as $$\Delta$$ U = q + w, where $$\Delta U$$ internal is change in internal energy, q is heat absorbed and w is work done.
Question 7
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For a reaction to be spontaneous at any temperature, the conditions are:

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

$$\Delta G = \Delta H - T\Delta S$$
If $$\Delta H$$ = negative and $$\Delta S$$ is positive , then for any value of T , $$\Delta G$$ will be negative and the process will be spontaneous.
Question 8
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Which of the following statements is not correct?

A
For a spontaneous process, $$\Delta G$$ must be negative.

B
Enthalpy, entropy, free energy etc. are state variables.

C
A spontaneous process is reversible in nature.

D
Total of all possible kinds of energy of a system is called its internal energy.

Solution

For a spontaneous process, $$\Delta G<0$$
The variables which depend only on the initial and final states of the system are state variables.
Therefore, enthalpy, entropy and free energy are state variables.
A spontaneous process is not reversible in nature as for opposite reaction to occur, a force has to be applied.
The internal energy is the total of all possible kinds of energy in the system.
Question 9
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A reaction proceeds through two paths I and II to convert $$X$$ $$\rightarrow$$ $$Z$$. What is the correct relationship between $$Q$$, $$Q_1$$ and $$Q_2$$ ($$Q$$ represents a change in internal energy, here) ?

A
$$Q = Q_1 \times Q_2$$

B
$$Q = Q_1 + Q_2$$

C
$$Q = Q_2 - Q_1$$

D
$$Q = Q_1 / Q_2$$

Solution

Internal energy (Q) is a state function.

$$\therefore { Q }_{ XZ }={ Q }_{ XY }+{ Q }_{ YZ }$$

$$ \therefore Q={ Q }_{ 1 }+{ Q }_{ 2 }$$

Hence, the correct answer is option $$\text{B}$$.
Question 10
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According to the first law of thermodynamics, $$\Delta U = q + w$$. In special cases the statement can be expressed in different ways. Which of the following is not a correct expression?

A
At constant temperature: $$q = - w$$

B
When no work is done: $$\vartriangle U = q$$

C
In gaseous system: $$\Delta U = q + P\Delta V$$

D
When work is done by the system: $$\Delta U = q + w$$

Solution

At constant temperature, $$\Delta T=0$$
$$\therefore \Delta U=n{ C }_{ v }\Delta T\\ \therefore \Delta U=q+w=0\\ q=-w$$
When work done is zero, $$w=0$$
$$\therefore \Delta U=q+w\\ \therefore \Delta U=q$$
In gaseous system, $$w=P\Delta V$$
$$\therefore \Delta U=q+w\\ \therefore \Delta U=q+P\Delta V$$
When work is done by the system, w is negative.
$$\therefore \Delta U=q-w$$

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Thermodynamics Test - 34

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Thermodynamics Test - 34

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Question 1

$$\Delta$$ H is positive, $$\Delta$$ S is positive

$$\Delta$$ H is positive, $$\Delta$$ S is negative

$$\Delta$$ H is negative, $$\Delta$$ S is negative

$$\Delta$$ H is negative, $$\Delta$$ S is positive

Solution

Question 2

$$\Delta$$G will be positive at low temperature hence reaction is spontaneous at low temperature.

$$\Delta$$G will be negative at high temperature hence reaction is spontaneous at high temperature.

$$\Delta$$G will be negative at low temperature hence reaction is spontaneous at low temperature.

$$\Delta$$G will be negative at all temperature hence reaction is spontaneous at all temperatures.

Solution

Question 3

transition

formation

reaction

all of these.

Solution

Question 4

If $$\Delta$$G is negative (< 0), the process is non-spontaneous.

If $$\Delta$$G is positive (> 0), the process is non-spontaneous.

If $$\Delta$$G is negative (< 0), the process is spontaneous.

If $$\Delta$$G is positive (> 0), the process is equilibrium.

Solution

Question 5

$$\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} > 0$$

$$\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} < 0$$

$$\Delta S_{system} = \Delta S_{total} + \Delta S_{surr} > 0$$

$$\Delta S_{surr} = \Delta S_{total} + \Delta S_{system} < 0$$

Solution

Question 6

first law of thermodynamics

second law of thermodynamics

third law of thermodynamics

zeroth law of thermodynamics

Solution

Question 7

$$\Delta$$ 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 8

For a spontaneous process, $$\Delta G$$ must be negative.

Enthalpy, entropy, free energy etc. are state variables.

A spontaneous process is reversible in nature.

Total of all possible kinds of energy of a system is called its internal energy.

Solution

Question 9

$$Q = Q_1 \times Q_2$$

$$Q = Q_1 + Q_2$$

$$Q = Q_2 - Q_1$$

$$Q = Q_1 / Q_2$$

Solution

Question 10

At constant temperature: $$q = - w$$

When no work is done: $$\vartriangle U = q$$

In gaseous system: $$\Delta U = q + P\Delta V$$

When work is done by the system: $$\Delta U = q + w$$

Solution

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