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. The sublimation energy of I2(s) is 57.3 kJ/mol and the enthalpy of fusion is 15.5 kJ/mol. The enthalpy of vaporisation of I2 is
Solution
(a) According to Hess’s law total heat changes during a chemical reaction are independent of path of reaction. < br> Given, I2(s)→I2(g),∆H1=57.3 kJ/mol …(i) I2(s)→I2(l),∆H2=+15.5 kJ/mol …(ii)
Required equation I2(l)→I2(g),∆H1=? subtract Eq. (ii) from Eq. (i)
∴I2(l)→I2(g), ∆H=57.3+(-15.5) =+41.8 kJ/mol
(a) According to Hess’s law total heat changes during a chemical reaction are independent of path of reaction. < br> Given, I2(s)→I2(g),∆H1=57.3 kJ/mol …(i) I2(s)→I2(l),∆H2=+15.5 kJ/mol …(ii)
Required equation I2(l)→I2(g),∆H1=? subtract Eq. (ii) from Eq. (i)
∴I2(l)→I2(g), ∆H=57.3+(-15.5) =+41.8 kJ/mol
Q2.The heat of formations of CO(g) and CO
Solution
(a) C(s)+1/2 O2 (g)→CO2 (g);∆H1=-26.4
C(s)+1/2 O2 (g)→CO2 (g);∆H_2=-96.0 We have to find for ∆H, for reaction
CO+1/2 O2→CO2;∆H1=? Eqs.(ii)-(i), we get
∆H=-94-(-26.4)=-67.6 kcal
(a) C(s)+1/2 O2 (g)→CO2 (g);∆H1=-26.4
C(s)+1/2 O2 (g)→CO2 (g);∆H_2=-96.0 We have to find for ∆H, for reaction
CO+1/2 O2→CO2;∆H1=? Eqs.(ii)-(i), we get
∆H=-94-(-26.4)=-67.6 kcal
Q3. When a bottle of perfume is opened, odourous molecules mix with air and slowly diffuse throughout
the entire room. The incorrect fact about the process is;
the entire room. The incorrect fact about the process is;
Solution
d) The randomness in gaseous state is more than liquid state.
d) The randomness in gaseous state is more than liquid state.
Q4. In which reaction ∆S is positive?
Solution
c) SV°-SL°=+ve. Conversion of liquid to vapour occurs process.
c) SV°-SL°=+ve. Conversion of liquid to vapour occurs process.
Q5.In an isothermal process
Solution
b) For an isothermal process ∆T=0 and ∆E=0 and q≠0
b) For an isothermal process ∆T=0 and ∆E=0 and q≠0
Q6. Work done in reversible isothermal process is given by:
Solution
a) This is the derived formula for W_rev is isothermal change.
a) This is the derived formula for W_rev is isothermal change.
Q7.Which are correct to express work terms?
Solution
(d) These are the expressions to explain work.
(d) These are the expressions to explain work.
Q8.A gas expands isothermally and reversibly. The work done by the gas is:
Solution
(c) Maximum work is obtained under reversible conditions (either isothermal or adiabatic).
(c) Maximum work is obtained under reversible conditions (either isothermal or adiabatic).
Q9. In an isobaric process, the ratio of heat supplied to the system (dθ) and work done by the system (dW) for diatomic gas is
Solution
(b) -dQ/dW=dQ/(dQ-dE);[∵dE=dW+dQ:-dW=dQ-dE] =(nCpdT)/(nCp dT-nCvdT)=Cp/((Cp-Cv)) =7/2 for diatomic gas
(b) -dQ/dW=dQ/(dQ-dE);[∵dE=dW+dQ:-dW=dQ-dE] =(nCpdT)/(nCp dT-nCvdT)=Cp/((Cp-Cv)) =7/2 for diatomic gas
Q10. Thermodynamics is concerned with:
Solution
b)
b)