Thermodynamics Test

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Thermodynamics Test - 81

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Weekly Quiz Competition

Question 1
1 / -0

Hess's law of constant heat summation in based on:

A
$$E=mc^{2}$$

B
Conservation of mass

C
First law of thermodynamics

D
None of the above

Solution

Hess’s law is an application of first law of thermodynamics.
Option C is correct.
Question 2
1 / -0

Hess law of heat summation includes

A
Initial reactants only

B
Initial reactants and final products

C
Final products only

D
Intermediates only

Solution

Hess law includes initial reactants and final products.
Option B is correct.
Question 3
1 / -0

In the reaction for the transition of carbon in the diamond form to carbon in the graphite form, $$\Delta H$$ is -453.5 cal. This points out that

A
graphite is chemically different from diamond

B
graphite is as stable as diamond

C
graphite is more stable than diamond

D
diamond is more stable than graphite

Solution

Kinetically diamond is more stable than graphite but thermodynamically graphite is more stable than diamond. The thermodynamic stability of a compound plays an important role in the stability of a compound.
So, graphite is more stable than diamond.
Hence, Option "C" is the correct answer.
Question 4
1 / -0

Gibb's free energy (G) is defined as:

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

B
$$\Delta G=\Delta H+\frac{T}{\Delta S}$$

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

D
$$\Delta G=\Delta H+T\times C_{P}$$

Solution

Gibbs energy is showing $$\Delta G= \Delta H -T \Delta S$$
Question 5
1 / -0

Conditions of standard state used in thermochemistry is?

A
$$0^{\circ}C$$ and 1 atm

B
$$20^{\circ}C$$ and 1 atm

C
$$25^{\circ}C$$ and 1 atm

D
0 K and 1 atm

Solution

Condition of standard state used in $$25°C and 1 atm
Question 6
1 / -0

$$C+O_{2}\rightarrow CO_{2}+94.2 kcal$$
$$H_{2}+1/2O_{2}\rightarrow H_{2}O+68.3 kcal$$
$$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O\left ( g \right
)+210.8kcal$$
Then the possible heat of formation of methane will be

A
47.3 kcal

B
20.0 kcal

C
45.9 kcal

D
-47.3 kcal

Solution

Given:
$$C+O_{2}\rightarrow CO_{2}; \ \Delta H_f (CO_2) = +94.2 kcal$$ (i)

$$H_{2}+1/2O_{2}\rightarrow H_{2}O; \ \Delta H_f(H_2O) =+68.3 kcal$$ (ii)

$$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O\left ( g \right
); \ \Delta H_R=+210.8kcal$$ (iii)
$$\Delta H_R$$ for reaction (iii) can be written as:
$$\Delta H = \Delta H_f (CO_2) +2\times \Delta H_f (H_2O) - \Delta H_F (CH_4) -2\times \Delta H_f (O_2)$$
$$\implies 210.8 = 94.2 + 2 \times 68.3 -\Delta H_f (CH_4) - 2\times 0$$
$$\Delta H_f (CH_4) = 20 \ kcal$$
Hence, Option "B" is the correct answer.
Question 7
1 / -0

Which of the following is slow process

A
Isothermal

B
Adiabatic

C
Isobaric

D
None of these

Solution

isothermal processes are necessarily slow as they require heat transfer to remain at the same temperature which is done by being in thermal equilibrium with some reservoir. A process will be isothermal only if it happens on timescales larger than the timescale required for effective heat transfer.
Question 8
1 / -0

In the Born-Haber cycle for the formation of solid common salt (NaCl), the largest contribution comes from :

A
the low ionization potential of Na

B
the high electron affinity of Cl

C
the low $$\triangle H_{vap}$$ of Na(S)

D
the lattice energy

Solution
Question 9
1 / -0

If enthalpy of an overall reaction $$X \rightarrow Y$$ along one rout is $$\Delta H$$ and $$\Delta H_{1}, \Delta H_{2}, \Delta H_{3},....$$ representing enthalpies of reactions leading to same product Y then $$\Delta H$$ is

A
$$\Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3}...$$

B
$$\Delta H = \Delta H_{1} \times \Delta H_{2} \times \Delta H_{3}...$$

C
$$\Delta H = \Delta H_{1} + \Delta H_{2} - \Delta H_{3}....$$

D
$$\Delta H = \dfrac{\Delta H_{1} \times \Delta H_{2} \times \Delta H_{3}}{2}....$$

Solution

According to Hess's law,

$$ \Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3}......$$

Where $$ \Delta H_{1}, \Delta H_{2} , \Delta H_{3} ,$$ etc. are enthalpies of various steps leading to final same product.
Question 10
1 / -0

The statement "The change of enthalpy of a chemical reaction is same whether the reaction takes place in one or several steps" is

A
Le Chatelier's law

B
van't Hoff's law

C
first law of thermodynamics

D
Hess's law.

Solution

The statement "The change of enthalpy of a chemical reaction is same whether the reaction takes place in one or several steps" is $$\text{Hess's law of constant heat summation.}$$

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Thermodynamics Test - 81

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Thermodynamics Test - 81

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Question 1

$$E=mc^{2}$$

Conservation of mass

First law of thermodynamics

None of the above

Solution

Question 2

Initial reactants only

Initial reactants and final products

Final products only

Intermediates only

Solution

Question 3

graphite is chemically different from diamond

graphite is as stable as diamond

graphite is more stable than diamond

diamond is more stable than graphite

Solution

Question 4

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

$$\Delta G=\Delta H+\frac{T}{\Delta S}$$

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

$$\Delta G=\Delta H+T\times C_{P}$$

Solution

Question 5

$$0^{\circ}C$$ and 1 atm

$$20^{\circ}C$$ and 1 atm

$$25^{\circ}C$$ and 1 atm

0 K and 1 atm

Solution

Question 6

47.3 kcal

20.0 kcal

45.9 kcal

-47.3 kcal

Solution

Question 7

Isothermal

Adiabatic

Isobaric

None of these

Solution

Question 8

the low ionization potential of Na

the high electron affinity of Cl

the low $$\triangle H_{vap}$$ of Na(S)

the lattice energy

Solution

Question 9

$$\Delta H = \Delta H_{1} + \Delta H_{2} + \Delta H_{3}...$$

$$\Delta H = \Delta H_{1} \times \Delta H_{2} \times \Delta H_{3}...$$

$$\Delta H = \Delta H_{1} + \Delta H_{2} - \Delta H_{3}....$$

$$\Delta H = \dfrac{\Delta H_{1} \times \Delta H_{2} \times \Delta H_{3}}{2}....$$

Solution

Question 10

Le Chatelier's law

van't Hoff's law

first law of thermodynamics

Hess's law.

Solution

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