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Quiz 186
Mdcat-past-papers Quiz
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3701. The sum of all the energies of all the molecules or atoms of a substance is called its:
Specific heat
Heat capacity
Latent heat
Internal energy
3702. If the work is done by the system on the surroundings and heat is absorbed by the system, the change in internal energy ΔE will be:
ΔE = q + w
ΔE = q +(– w)
ΔE = –q – w
ΔE = –q + w
3703. Calculate the work done when 1 mole of an ideal gas expands from 15 dm³ to 20 dm³ against constant external pressure of 2 atmospheres.
-10atm dm3
-5 atm dm3
-20 atm dm3
10atm dm3
3704. One slice of bread with a tablespoon of peanut butter on it contains 20g carbohydrate, 10g protein, and 9g fat. Calculate total energy consumed in this intake:
158 kcal
201 kcal
173 kcal
218 kcal
3705. One calorie is equal to:
4.18 J
4.18 J
0.418 kJmol⁻¹
0.418 kJ
3706. According to Hess's law, which one is a state function?
Work
Heat
Both are state function
Both are not state function
3707. First law of thermodynamics is represented as:
ΔH = q + w
ΔE = q + w
ΔH = ΔE + W
ΔH = ΔE + q
3708. Expansion takes place when a gas is evolved during a chemical reaction between marble chips and dilute HCl. Predict the energy change and work done:
w is positive, q is negative
w is negative, q is positive
q is positive, no work done
w is positive, q is positive
3709. In standard enthalpy of atomization, heat of surrounding:
Remains unchanged
Increases
Increases then decreases
Decreases
3710. ΔH will be given a negative sign in:
Exothermic reaction
Dissociation reaction
Decomposition reaction
Endothermic reaction
3711. Lattice energy of an ionic crystal is the enthalpy of:
Combustion
Dissociation
Dissolution
Formation
3712. Heat of formation (ΔH°f) for CO₂ is:
–394 kJ/mol
394 kJ/mol
–294 kJ/mol
–390 kJ/mol
3713. 2H₂O → ΔH = +285.5 kJ/mol What will be the enthalpy change in the equation?
285.5 kJ/mol
Zero kJ/mol
–285.5 kJ/mol
1 kJ/mol
3714. Combustion of graphite to form CO₂ can be done by two ways. Reactions are: C + O₂ → CO₂ ΔH = –284 kJ/mol C + ½O₂ → CO ΔH = –110 kJ/mol CO + ½O₂ → CO₂ ΔH = –676 kJ/mol What will be enthalpy of formation of CO?
–110 kJ/mol
–110 kJ/mol
110 kJ/mol
676 kJ/mol
3715. Reaction of water with quick lime results in the rise of temperature of the system. Using the concept of energy change, indicate the nature of the reaction:
Third-order reaction
Non-spontaneous reaction
Endothermic reaction
Exothermic reaction
3716. The equation that represents enthalpy of atomization of hydrogen is:
½H₂O(g) → H(g) +218 kJ mol⁻¹
½H₂(g) → H(g) +218 kJ mol⁻¹
H₂O(g) → 2H(g) +218 kJ mol⁻¹
½H₂(g) → H(g) –218 kJ mol⁻¹
3717. Standard enthalpy of combustion of graphite at 25°C is –395.41 kJ/mol and that of diamond is –393.5 kJ/mol. The enthalpy change of graphite to diamond is:
–1.91
2.1
2.91
1.91
3718. ½ H2 (g) →H(g) ΔH =218 kJ/mol in this reaction ΔH will be called
Anthalpy of atomization
Anthalpy of above de-composition
Anthalp of formation
Anthalpy of association
3719. Mg + ½O₂ → MgO(g) ΔH = –692 kJ/mol at STP. Enthalpy of above reaction will be called:
ΔH⁰ₐₜ
ΔH⁰ₛ
ΔH⁰ₛₒₗ
ΔH⁰f
3720. Which one of the following enthalpy changes is always exothermic?
Enthalpy of combustion
Enthalpy of solution
Enthalpy of atomization
Enthalpy of fusion
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