Thermodynamics Test

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Thermodynamics Test - 4

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Question 1
4 / -1

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound

A
is always negative

B
is always positive

C
may be positive or negative

D
is never negative

Solution

Depend on whether the formation of compounds is exothermic or endothermic.For exothermic reaction, enthalpy of formation would be negative and for endothermic,it is positive.
Question 2
4 / -1

Given,

A

B

C

D

Solution

Applying i-ii/2-3/2iii
We get ∆_fH of NCl₃ in terms of ∆H1, ∆H₂ and ∆H₃
Question 3
4 / -1

Diborane is a potential rocket fuel which undergoes combustion according to the reaction,
From the following data, enthalpy change for the combustion of diborane is

A
-36 kJ mol^-1

B
-1237 kJ mol^-1

C
-2035 kJ mol^-1

D
-1639 kJ mol^-1

Solution

On applying equation 1 + 3 times equation 2 minus equation 4 plus equation 3, we will get enthalpy change for the combustion of diborane.
Question 4
4 / -1

Following diagram represents Born-Haber cycle to determine lattice energy of NaCI(s). It is based on Hess’s law of constant heat summation. Δ_latticeH° of

A
-788 kJ mol^-1

B
+788 kJ mol^-1

C
-411.2 kJ mol^-1

D
+411.2 kJ mol^-1

Solution

The correct answer is Option A.

-411.2 = 108.4 + 495.6 + 121.0 - 348.6 + LE
=>LE= -411.2 - 108.4 - 495.6 - 121.0 +348.6
=> LE = -787.6 ~ -788 kJ mol^-1
Question 5
4 / -1

Thus,

A
Q = S + I + D - E - U

B
Q = S + 2I + D - E - U

C
Q = S + 1 + D - 2E - U

D
Q = S +1 + D - E + U

Solution

Put the things we need as reactant on LHS the product on RHS. To do that, we will apply equation S+I+D-2E-U.
Question 6
4 / -1

Given, HCI (g) → H (g) + Cl (g) at 298 K (say temperature T K),
For
Q. Thus, for the given reaction (at 0 K) is

A
428.136 kJ

B
-428.136 kJ

C
431.961 kJ

D
-431.961 kJ

Solution

Ho= Ho of product - Ho of reactant
so ho =(6.197 + 6.272 ) - 8.644
= 3.825
Ho298-Ho = 431.961 - 3.825=428.136532873
Question 7
4 / -1

The measured enthalpy change for burning of ketene (g) (CH₂CO) is - 981.1 kJ mo^-1 at 298 K ,CH₂CO (g)+ 2O₂(g) → 2CO₂ (g) + H₂O (g) and that of CH₄ (g) is - 802.3 kJ mol^-1 at 298 K
Thus, enthalpy change at 298 K for the following thermochemical reaction is

A
-623.5 kJ

B
+623.5 kJ

C
-178.8 kJ

D
+802.3 kJ

Solution

CH₂CO (g)+ 2O₂(g) → 2CO₂ (g) + H₂O (g) -(i)
CH4 (g)+ 2O₂(g) → CO₂ (g) + 2H₂O (g) -(ii)
On 2(ii)-(i)
2CH₄ (g)+ 2O₂(g) → CH₂CO (g) + 3H₂O (g) -(iii)
∆H_(iii) = 2∆H_(ii) - ∆H_(i) = [2 (-802.3) -(-981.1)] -623.5 kJ mol^-1
Question 8
4 / -1

Direction (Q. Nos. 8) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q. In the following thermochemical reactions,

Statement I : Enthalpy of combustion of CO(g) is - 26 kcal mol^-1.
Statement II : CO₂(g) is an exothermic compound.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

According to me, the question is presented in an improper way. For statement I to be correct, value of enthalpy of combustion of CO be -68 kcal mol^-1. This can be sensed by reversing reaction II adding it to reaction I. By this, we will get the value of enthalpy of combustion of CO be -68 kcal mol^-1.
However, statement II is wrong. There is no such exothermic or endothermic compounds. Reactions are exothermic and not the products.
Question 9
4 / -1

Direction (Q. Nos. 9 and 10) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (a),(b),(c),(d).
Consider the following thermochemical equations

Q. Δ_rH° of the following thermochemical reaction is
CIO₂ (g) + O (g) → CIO₃ (g)

A
-197 kJ

B
+197 kJ

C
+312 kJ

D
+212 kJ

Solution

The question can be simply done by understanding that this reaction is endothermic so ∆H is negative.
Hence A is correct.
Question 10
4 / -1

Consider the following thermochemical equations

Q. Δ_rH° of the following thermochemical reaction is

A
107.0 kJ

B
-91.5 kJ

C
-312.0 kJ

D
-196.0 kJ
Question 11
4 / -1

Direction (Q. Nos. 11) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Q. For liquid water at

Compare the parameters ΔH₁, ΔH₂ ... in Column I with the corresponding values in kJ in Column II.

A
a

B
b

C
c

D
d

Solution

∆H¹=∆rH298=-285.83kJmol-¹
∆H²=Cp(liq)∆T=75.29×75 =5.64kJmol-¹
∆H³=∆Hvap =40.88kJmol-¹
∆H⁴=Cp(gas)∆T=35.57×(-75) =-2.5183kJmol-¹
Question 12
4 / -1

The internal energy of a perfect gas is :

A
completely kinetic

B
completely potential

C
sum of potential and kinetic energy of the molecules

D
difference of kinetic and potential energy of the molecules

Solution

In a perfect gas or an ideal gas, the atoms of molecules collide in a perfectly elastic manner. That is the intermolecular attractive forces are absent and thus the internal energy is completely kinetic.

Hence, option A is correct.
Question 13
4 / -1

Energy hidden in a definite quantity of substance is:

A
Enthalpy

B
Internal energy

C
Free energy

D
Entropy

Solution

Internal energy is the hidden energy in a definite quantity of substance
Question 14
4 / -1

Which of the following relations is not correct?

A
G = H − TS

B

C

D

Solution
Question 15
4 / -1

Hess's law of constant heat summation is based on

A
E = mc²

B
conservation of mass

C
first law of thermodynamics

D
E = hν

Solution

Hint: Hess's law of constant heat summation states that " Regardless of multiple stages, total enthalpy changes is the sum of all the changes."

Correct answer: Option C

Explanation for Correct Option:

Mathematically Hess law can be represented as;