The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry..
Q1. Given,
NH3(g)+3Cl2(g) ⇌NCl3(g)+3HCl(g); - ∆H1
N2(g)+3H2(g)⇌2NH3(g); -∆H2 H2(g)+Cl2(g)⇌2HCl(g);∆H3
The heat of formation of NCl3(g) in the terms of ∆H1,∆H2 and ∆H3 is:
N2(g)+3H2(g)⇌2NH3(g); -∆H2 H2(g)+Cl2(g)⇌2HCl(g);∆H3
The heat of formation of NCl3(g) in the terms of ∆H1,∆H2 and ∆H3 is:
Solution
(a) Find 1/2 N2+3/2 Cl2 ⟶NCl3; Multiply Eqs. (ii) by 1/2, (iii) 3/2 and subtract from Eq.(i); we get ∆Hf= -∆H1- [-((∆H2)/2 +3/2 ∆H3)]
= - ∆H1+(∆H2)/2 -3/2 ∆H3.
(a) Find 1/2 N2+3/2 Cl2 ⟶NCl3; Multiply Eqs. (ii) by 1/2, (iii) 3/2 and subtract from Eq.(i); we get ∆Hf= -∆H1- [-((∆H2)/2 +3/2 ∆H3)]
= - ∆H1+(∆H2)/2 -3/2 ∆H3.
Q2.Which has the least entropy?
Solution
b) Graphite possesses sp^2-hybridisation and has flat layer structure whereas diamond possesses sp^3-hybridisation and has rigid tetrahedral nature
b) Graphite possesses sp^2-hybridisation and has flat layer structure whereas diamond possesses sp^3-hybridisation and has rigid tetrahedral nature
Q3. The enthalpy change for the process, C(s) ⟶ C(g) is known as enthalpy of :
Solution
d) The process involves conversion of 1 mole of C(s) to C(g),i.e., sublimation.
d) The process involves conversion of 1 mole of C(s) to C(g),i.e., sublimation.
Q4. A boiled egg show a/an …in entropy.
Solution
a) No doubt solidification shows a decrease in entropy but in egg proteins structure are disordered in solid state due to denaturation
a) No doubt solidification shows a decrease in entropy but in egg proteins structure are disordered in solid state due to denaturation
Q5.Heat of neutralisation of which acid-base reaction is - 57.32 kJ for?
Solution
d) Strong acid (HNO3) and strong base (LiOH).
d) Strong acid (HNO3) and strong base (LiOH).
Q6. For a reaction, ∆H=9.08 kJ mol-1 and ∆S=35.7 JK-1 mol-1. Which of the following statement is correct for the reaction?
Solution
c) ∆H and ∆S both are +ve for spontaneous change and ∆H=+ve for endothermic reaction
c) ∆H and ∆S both are +ve for spontaneous change and ∆H=+ve for endothermic reaction
Q7. In a chemical reaction ∆H=150 kJ and ∆S=100 JK-1 at 300 K. The ∆G for the reaction is:
Solution
d) ∆G=∆H-T∆S ∆G=150-(100×300)/10-3) =120 kJ
d) ∆G=∆H-T∆S ∆G=150-(100×300)/10-3) =120 kJ
Q8.Enthalpy of vaporisation for water is 186.5 kJ mol-1. The entropy change during vaorisation is …kJ K-1 mol-1
Solution
a) ∆S=(∆Hv)/T ∆H=186.5 kJ T=373 K ∴∆S=186.5/373=0.5 kJ K-1 mol-1
a) ∆S=(∆Hv)/T ∆H=186.5 kJ T=373 K ∴∆S=186.5/373=0.5 kJ K-1 mol-1
Q9. When water is added to quick lime, the reaction is :
Solution
c) CaO+H2O ⟶Ca(OH)2; ∆H= -ve; the solution of lime heats up.
c) CaO+H2O ⟶Ca(OH)2; ∆H= -ve; the solution of lime heats up.
Q10. The work done by a system is 8 J, when 40 J heat is supplied to it. The change in
internal energy of the system during the process is:
internal energy of the system during the process is:
Solution
a) q=∆U-W;-W is work done by the system ∆U=40-8=32 J (∵-W=8)
a) q=∆U-W;-W is work done by the system ∆U=40-8=32 J (∵-W=8)