## CHEMISTRY THERMODYNAMICS QUIZ-6

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.
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Q1.Standard molar enthalpy of formation of CO2 is equal to:
•  Zero
•  The standard molar enthalpy of combustion of gaseous carbon
•  The sum of standard molar enthalpies of formation of CO2 and O2
•  The standard molar enthalpy of combustion of carbon (graphite)
Solution
Standard molar enthalpy of formation CO2 and the standard
molar enthalpy of combustion of carbon (graphite) refer to the
same chemical equation:
C(graphite)+O2(g) → CO2(g)

Q2.The molar enthalpies of combustion of C2H2(g) , C(graphite) and H2(g) are -1300, -394, and -286 kJ mol-1, respectively. The standard enthalpy of formation of C2H2(g) is
•  -226 kJ mol-1
•  -626 kJ mol-1
•   226 kJ mol-1
•   626 kJ mol-1
Solution
226 kJ mol-1
Q3.The standard heat of combustion of Al is-837.8 kJ mol-1at 25℃. If Al reacts with O2 at 25℃, which of the following release 250 kJ of heat?
•  The reaction of 0.624 mol of Al
•  The formation of 0.624 mol of Al2O3
•  The formation of 0.312 mol of Al
•  The formation of 0.150 mol of Al2O3
Solution

0.624 mol of Al = 837.8×0.624 on combustion gave=523 kJ
Hence false
Formation of 0.624 mol of Al2O3 gave
= 837.8×2×0.624
= 1045 kJ
Hence, false
0.312 mol of Al on combustion gave = 261 kJ. Hence, false
Formation of 0.150 mol of Al2O3 gave = 251.3 kJ. Hence, true

Q4.The enthalpy of formation of hypothetical MgCl is -125 kJ mol-1 and for MgCl2 is-642 kJ mol-1. What is the enthalpy of the disproportionation of MgCl
•   392 kJ mol-1
•  -392 kJ mol-1
•  -767 kJ mol-1
•  -517 kJ mol-1
Solution
Mgs + Cl2(g) ⟶ MgCl2(s) ,∆H1 = -642
Mgs + 12Cl2(g) ⟶ MgCls ,∆H2 = -125
2MgCl ⟶ MgCl2 + Mg ,∆H = ?
∆H = ∆H1 - 2∆H2 = -642 - 2×(-125) = -392 kJ mol-1

Q5.At equilibrium state
•  totalS > 0
•  totalS < 0
•  totalS = 0
•  Unpredictable
Solution
totalS = 0

Q6.The relationship between enthalpy and internal energy change is
•  ∆U = ∆H + P∆V
•  ∆H = ∆U + P∆V
• ∆H = ∆U - P∆V
•  P∆V = ∆U + ∆H
Solution
∆H = ∆U + P∆V

Q7.Enthalpy change of a reaction will be equal to
•  ∆U + P∆V
•  ∆U + V∆P
•  ∆U + ∆(PV)
•  ∆U + (∆ng)∆(PV)
Solution
∆H = ∆U + ∆(PV)
= ∆U + ∆P∆V

Q8.For the given reactions
SiO2 + 4HF → SiF4 + 2H2O , ∆H = -10.17 kcal
SiO2 + 4HCl → SiCl4 + 2H2O , ∆H = 36.7 kcal
It may be concluded that
•  HF will attack SiO2 and HCl will not
•  HCl will attack SiO2 and HF will not
•  HF and HCl both attack SiO2
•  None attack SiO2
Solution
∆G = ∆H - T∆S = -ve
All exothermic reactions are spontaneous, hence HF will attack SiO2

Q9.The word ‘standard’ in standard molar enthalpy change implies
•  Temperature 298 K
•  Pressure 1 atm
•  Temperature 298 K and pressure 1 atm
•  All temperatures and all pressures
Solution
Standard state implies to 1 atm pressure and 298 K temperature

Q10. Which law of thermodynamic, introduces the concept of entropy
•  First law
•  Zeroth Law
•  Third Law
•  Second Law
Solution
Second Law

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET & IIT JEE COACHING: THERMODYNAMICS Quiz-6
THERMODYNAMICS Quiz-6