Thermodynamics Test

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Thermodynamics Test - 17

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Question 1
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Direction (Q. Nos. 1-7) This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Given,
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
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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
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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
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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
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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
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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
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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
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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
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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