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NEET Exam  >  NEET Test  >  Topic-wise MCQ Tests  >  Test: Hess's Law of Constant Heat Summation - NEET MCQ

Hess's Law of Constant Heat Summation - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Hess's Law of Constant Heat Summation (15 Questions)

You can prepare effectively for NEET Topic-wise MCQ Tests for NEET with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Hess's Law of Constant Heat Summation". These 15 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

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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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: Question 15



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