Wednesday, August 25, 2021

Chemical Thermodynamics: Chemistry

 

Q.1.  In an adiabatic expansion of ideal gas 

A) W= -triangle E

B) W = triangleE

C) triangleE = 0

D) W = 0

Explanation- 

incrementE = q - W            ( for expansion )

In case of adiabatic expansion q = 0

therefore space minus increment straight E equals straight W 


Q.2.  For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter , triangleU and W correspond to  

A) triangleU < 0 , W = 0

B) triangleU < 0 , W < 0

C) triangleU > 0 , W = 0

D) triangleU > 0 , W > 0

Explanation-

Since the reaction is carried out in closed vessel , ( incrementV = 0 ) so no work of expansion or contraction can be there . 

Hence by 1 to the power of st  law of thermodynamics incrementU = q 

The reaction is a redox reaction , therefore it must be exothermic i.e., q < 0

So that incrementU < 0 .


Q.3.  The bond dissociation energy of gaseous H subscript 2 comma C I subscript 2  and HCI are 104,58 and 103 kcal/mol respectively. Enthalpy of formation of HCI gas is 

A) 59 kcal

B) -265 kcal

C) -22 kcal

D) -29.5 kcal

Explanation-

Formation of HCI can be given as 

1 half straight H subscript 2 space plus space 1 half CI subscript 2 space space rightwards arrow space space HCI
stack straight B. straight E. with kcal below space colon space 1 half cross times 104 space space space space space 1 half cross times 58 space space space space space 103
increment straight H subscript Reaction space space end subscript straight i. straight e. comma space increment straight H subscript straight f left parenthesis HCI right parenthesis end subscript equals begin inline style sum for blank of end style space straight B. straight E. subscript left parenthesis Products right parenthesis end subscript
equals 52 plus 29 minus 103 equals negative 22 space kcal space mol to the power of negative 1 end exponent 


Q.4.  In an endothermic reaction , the value of triangle H  is 

A) Zero

B) Positive

C) Negative

D) Constant 

Explanation-

increment straight H equals straight H subscript straight p minus straight H subscript straight R
In space endothermic space reaction space straight H subscript straight p greater than straight H subscript straight R space comma space hence space increment straight H space is space positive space. 


Q.5.  2 mol of an ideal gas at 27degree C is expanded reversibly from 2 L to 20 L . Find entropy change ( R = 2 cal/ mol K ).

A) 92.1

B) 0

C) 4

D) 9.2


Q.6.  For melting of a solid at 25degree C , the fusion process requires energy equivalent to 2906 J to be added to system considering the process to be reversible at fusion point, the entropy change of the process is 

A) 9.75 J Kblank to the power of negative 1 end exponent molblank to the power of negative 1 end exponent

B) 11.272 J  Kblank to the power of negative 1 end exponent molblank to the power of negative 1 end exponent

C) 2.33 J Kblank to the power of negative 1 end exponent molblank to the power of negative 1 end exponent

D) Insufficient data 


Q.7. Compounds with high heat of formation are less stable because 

A) It is difficult to synthesis them 

B) Energy rich state leads to instability 

C) High temperature is required to synthesis them

D) Molecules of such compounds are disturbed 

Explanation-

More energy results in less stability.


Q.8.  Heat exchanged in a chemical reaction at constant temperature and pressure is called 

A) Entropy change

B) Enthalpy change

C) Internal energy change

D) Free energy change 

Explanation-

Enthalpy change is the heat exchanged at constant temperature and pressure 


Q.9.  In order to decompose 9 g of water , 142.5 kJ of heat is required. Hence enthalpy of formation of water is 

A) -142.5 kJ

B) +142.5 kJ

C) -285 kJ

D) +285 kJ


Q.10.  The value of triangle H subscript O minus H end subscript  is 109  kcal molblank to the power of negative 1 end exponent .Then formation of one mole of water in gaseous state from  Hblank subscript left parenthesis g right parenthesis end subscript  and Oblank subscript left parenthesis g right parenthesis end subscript is  accompanied by 

A) Release of 218 kcal of energy 

B) Release of 109 kcal of energy 

C) Absorption of 218 kcal of energy 

D) Unpredictable 

Explanation-

straight H subscript 2 straight O forms two O - H bonds . 

therefore increment straight H subscript straight f open parentheses straight H subscript 2 straight O close parentheses end subscript equals 2 space increment straight H subscript straight O minus straight H end subscript equals 2 cross times open parentheses 109 close parentheses space equals space 218 space kcal space mol to the power of negative 1 end exponent
Bond space formation space involves space release space of space energy space. space 






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