CHEMISTRY THERMODYNAMICS QUIZ-3
Dear Readers,
IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies.
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..
Q1.Which is not intensive property?:
Solution
Volume
Volume
Q2.Which one of the following statements is false?
Solution
Work is a state function
Work is a state function
Q3.The ∆fH for CO2(g), CO(g), and H2O(g)are -393.5, -110.5, and-214.8 kJ mol-1, respectively.
The standard enthalpy change (in kJ mol-1) for the reaction
CO2(g)+H2(g)⟶CO(g)+H2O(g) is
The standard enthalpy change (in kJ mol-1) for the reaction
CO2(g)+H2(g)⟶CO(g)+H2O(g) is
Solution
CO2(g)+H2(g)⟶CO(g)+H2O(g)
∆H⊝={[∆fH⊝CO, (g)+∆fH⊝H2O, (g)] -∆fH⊝CO2, (g)}
= [-110.5+(-241.8(--393.5+0]
= +41.2 kJ
∆fH⊝H2, (g) = 0
∆fH⊝ = 0 in elementary state
CO2(g)+H2(g)⟶CO(g)+H2O(g)
∆H⊝={[∆fH⊝CO, (g)+∆fH⊝H2O, (g)] -∆fH⊝CO2, (g)}
= [-110.5+(-241.8(--393.5+0]
= +41.2 kJ
∆fH⊝H2, (g) = 0
∆fH⊝ = 0 in elementary state
Q4.If ∆fH⊝ of ICl(g), Cl(g), and I(g)) is 17.57, 121.34 and 106.96 J mol-1 respectively. Then bond dissociation energy of I-Cl bond is
Solution
210.73 J mol-1
210.73 J mol-1
Q5.The difference between the heats of reaction at constant pressure and constant volume for the reaction
2C6H6(l)+15O2⟶12CO2(g)+6H2O(l) at 25℃ in kJ is
Solution
2C6H6(l)+15O2⟶12CO2(g)+6H2O(l)
∆H = ∆E + ∆ngRT
∆H - ∆E = ∆ngRT
∆ng = 12 - 15 = -3 ∆H - ∆E = -3×8.314×298 = -7432.716 J
2C6H6(l)+15O2⟶12CO2(g)+6H2O(l)
∆H = ∆E + ∆ngRT
∆H - ∆E = ∆ngRT
∆ng = 12 - 15 = -3 ∆H - ∆E = -3×8.314×298 = -7432.716 J
Q6.The heat of hydrogenation of ethane is x1 and that of benzene is x2
Hence resonance energy of benzene is
Solution
Q7.The relationship between the free energy change (∆G) and entropy change (∆S) at constant temperature (T) is
Solution
∆G = ∆H - T∆S
∆G = ∆H - T∆S
Q8.Molar heat capacity of water in equilibrium with ice at constant pressure is
Solution
By definition CP,m=dqp/dT
For H2O(l)⇌H2O(s),
Temperature does not change if some heat is given to the system. Hence
CP,m=+ve/Zero=∞
By definition CP,m=dqp/dT
For H2O(l)⇌H2O(s),
Temperature does not change if some heat is given to the system. Hence
CP,m=+ve/Zero=∞
Q9.H2+ 1/2O2 → H2O; ∆H=-68 kcal
K + H2O + aq → KOH(aq)+1/2H2; ∆H⊝=-48 kcal
KOH + aq → KOH(aq);∆H⊝ = -14 kcal
From the above data, the standard heat of formation of KOH in kcal is
Solution
-68 -48 + 14
-68 -48 + 14
Q10. If a gas absorbs 200 J of heat and expands by 500 cm3 against a constant pressure of 2×105 N m-2, then the change in internal energy is
Solution
∆U = q - w = q - P∆V
=200-2×105×500×10-6
=100 J
∆U = q - w = q - P∆V
=200-2×105×500×10-6
=100 J
