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NEET Exam  >  NEET Tests  >  Chemistry Class 11  >  Test: Hess's Law of Constant Heat Summation - NEET MCQ

Test: Hess's Law of Constant Heat Summation - NEET MCQ


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15 Questions MCQ Test Chemistry Class 11 - Test: Hess's Law of Constant Heat Summation

Test: Hess's Law of Constant Heat Summation for NEET 2024 is part of Chemistry Class 11 preparation. The Test: Hess's Law of Constant Heat Summation questions and answers have been prepared according to the NEET exam syllabus.The Test: Hess's Law of Constant Heat Summation MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Hess's Law of Constant Heat Summation below.
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Test: Hess's Law of Constant Heat Summation - 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

Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 1

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.

Test: Hess's Law of Constant Heat Summation - Question 2
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Given,



Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 2


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

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Test: Hess's Law of Constant Heat Summation - 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 



Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 3

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

Test: Hess's Law of Constant Heat Summation - 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. ΔlatticeH° of 
Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 4

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

Test: Hess's Law of Constant Heat Summation - Question 5
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Thus,
Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 5

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.

Test: Hess's Law of Constant Heat Summation - 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
Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 6

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

Test: Hess's Law of Constant Heat Summation - 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
Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 7

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

Test: Hess's Law of Constant Heat Summation - Question 8
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The condition for free expansion of an ideal gas under adiabatic condition is 
Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 8

Explanation:

For free expansion of ideal gas,

Pext = 0

WpV = 0

For adiabatic process, 

q = 0

Conclusion:

Thus, the condition for free expansion of an ideal gas under adiabatic condition is q = 0, w = 0, ΔT = 0.

Test: Hess's Law of Constant Heat Summation - Question 9
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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. ΔrH° of the following thermochemical reaction is

CIO2 (g) + O (g) →  CIO3 (g)
Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 9

The question can be simply done by understanding that this reaction is endothermic so ∆H is negative.

Hence A is correct.

Test: Hess's Law of Constant Heat Summation - Question 10
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Consider the following thermochemical equations



 

Q. ΔrH° of the following thermochemical reaction is
Test: Hess's Law of Constant Heat Summation - Question 11
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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.
Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 11

∆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

*Answer can only contain numeric values
Test: Hess's Law of Constant Heat Summation - Question 12
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Direction (Q. Nos. 12-15) This section contains 4 questions. when worked out will result in an integer from 0 to 9 (both inclusive).

Q. Based on (BE) values, ΔrH° of the following reaction 298 K is

Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 12

Let the BE value of B---Br bond be x and B---Cl bond be y
BE of BBr3 is 3x, BCl3=3y , BBr2Cl=2x+y ,BCl2Br=2y+x
then ∆Hr=Hr(products)-Hr(reactants)=(2x+y+2y+x) - (3x+3y)=0

*Answer can only contain numeric values
Test: Hess's Law of Constant Heat Summation - Question 13
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Heat of combustion of CH2CO(g) is - 981.1 kJ mol-1 and that of CH4 (g) is - 802.3 kJat 298 K. If 1247.0 kJ of heat is released in the following change

Q. How many moles of CH4(g)are used?

Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 13

[2CH4 + 2O2 → CH2O + H2O]n
∆HC = n[-981.1-(-8023×2)]
1247       = n[-981.1+1604.6] 
N[623.5] = 1247
N = 4
So, no of moles of CH4 = n×2 = 2×2 = 4

*Answer can only contain numeric values
Test: Hess's Law of Constant Heat Summation - Question 14
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Consider the following reactions


Q. What is the resonance energy (in kcal) of A?

Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 14

Resonance energy = (Observed ∆H) - (Theoretical ∆H)
= (-98.0) - 2(-45.0)
= -8 kcal mol-1
So we have Resnance energy = 8

*Answer can only contain numeric values
Test: Hess's Law of Constant Heat Summation - Question 15
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Given, enthalpy of combustion of carbon (s) = -94 kcal mol-1

H2(g) = - 68 kcal mol-1 and CXH4 = - 318 kcal mol-1 and  ΔfH° (CxH4) = - 100 kcal mol-1

Q. What is the value of x?

Detailed Solution for Test: Hess's Law of Constant Heat Summation - Question 15



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