## Thermochemistry Quiz-12

Thermochemistry is the part of thermodynamics that studies the relationship between heat and chemical reactions. As we know thermodynamics is a very important topic from examination point of view. Hence as a part of it, thermochemistry also becomes important for exams. So if you want to be best in thermodynamics then you must study thermochemistry properly. So go ahead and prepare well. All the best !.

Q1. For a given substance T1 and T2 are freezing point and melting point of a substance. Which of the graph represents correctly, the variation of ΔS with temperature?
Solution:-
T1 and T2 are same for a substance.

•  Pressure is maintained constant
•  Gas is isothermally expanded
•  There is perfect heat insulation
•  The system changes heat with surroundings
Solution:-
An adiabatic process is one in which exchange of heat is not taking place in between system and surroundings. This can be made by putting insulation at the boundries of system

Q3. The work done by a system in an expansion against a constant external pressure is:
•  ΔP. ΔV
•  −P. ΔV
•  Q
•  V. ΔP
Solution:-
Wirr. for a process at constant pressure = −PΔV; note that work is irreversible if expansion is made at constant pressure.

Q4. The free energy change for a reversible reaction at equilibrium is
•  Large, positive
•  Small, negative
•  Small, positive
•  0
Solution
D

Q5. Which plot represents for an exothermic reaction?
Solution:-
As, ΔH = HP − HR
ΔH is negative hence, HP < HR.

Q6. Hess’s law is based on
•  Law of conservation of mass
•  Law of conservation of energy
•  First law of thermodynamics
•  None of the above
Solution:-
Hess’s law is based upon law of conservation of energy i.e., first law of thermodynamics.

Q7. Select the incorrect statement
•  P.V work is usually negligible for solid and liquid
•  For a closed system with P−V work only, an isobaric process that has q = +ve must have ΔT = +ve.
•  For a cyclic process q = 0
•  Black phosphorus is most stable form of P but H°f = 0 for white phosphorus.
Solution:-
Rest all are correct.

Q8. 48 g of C (diamond) on complete combustion evolves 1584 kJ of heat. The standard heat of formation of gaseous carbon is 725 kJ/mol. The energies required for the process
(i) C(graphite) → C(gas)
(ii) C(diamond) → C(gas) are:
•  725, 727
•  717, 725
•  725, 723
•  None of these
Solution
As graphite is thermodynamically more stable than diamond hence, more heat is required to convert graphite to gaseous carbon

Q9. An ideal gas expands in volume from 1 × 103 m3 to 1 × 102 m3 at 300 K against a constant pressure of 1 × 105 N m−2. The work done is
•  270 kJ
•  −900 kJ
•  −900 J
•  900 kJ
Solution:-
W = −PΔV = −1 × 105(1×10−2 − 1×10−3) = − 900 J

Q10. A mixture of hydrogen and chlorine on exposure to ultra violet sunlight reacts with explosion. The step involved in the initiation of the reaction is :
•  H2 → H + H
•  Cl + Cl → Cl2
•  H2 + Cl2 → 2HCl
•  Cl2 → Cl + Cl
Solution
It is Cl2 molecule which dissociates to give free radicals on exposure to light. ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: Thermochemistry Quiz-12
Thermochemistry Quiz-12