Question # 1

ΔH° represent the enthalpy change at

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0°C and 1 atm pressure

25°Cand 1atm

0K and I atm pressure

25°C and 2 atm pressure

The change in enthalpy when one mole of a substance is dissolved in a specified quantity of solvent at a given temperature is called

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Heat of reaction

Heat of solvation

Heat of combustion

Heat of solvent

NaOH+HCI- NaCI+ H2O. Enthalpy change in the above reaction is called

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Enthalpy of reaction

Enthalpy of Neutralisation

Enthalpy of formation

Enthalpy of combustion

Neutralization of acid-base is

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Spontaneous

Exothermic

Non spontaneous

Both "a" and "c

Enthalpy of neutralization (∆H°n) per mole of H2SO4/ Ba(OH)2 is

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+57.4 kJmol-1

-114.8 kJmol-1

-57.4 kJmol-1

Enthalpy of formation of one mole of ionic compound form gaseous ion under standard condition is called

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Gibb's energy

Bond energy

Lattice energy

If internal energy of the system is increased

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Change in state of the system may occur

Temperature of the system may rise

Chemical reaction may take place

All of these

Most of the reactions which give stable products are

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Endothermic

Exothermic

Isothermal

Non of these

The heat of reaction depends upon

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Temperature of the reactants

Physical states of the reactants and the products

Both A) and B)

Path of the reaction and the temperature

Enthalpy of a system can be calculated by which of following relationship

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q=ΔΕ

q=m×S×∆T

q=pv

q=m×v×∆T

If a reaction involves only solids and liquids, which of the following is true?

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ΔH = ΔE

ΔΗ =ΔΕ

ΔH>ΔΕ

AH =AE +nRT

What is correct about heat of combustion

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It is applicable to gaseous substances only

It is always negative

It is always positive

It is positive in some cases while negative in other

The enthalpy change AH of a process is given by the relation

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ΔΗ =ΔΕ + P∆V

ΔH=ΔE +W

ΔH =ΔΕ-ΔnRT

ΔΕ =ΔΗ + P∆V

If an endothermic reaction is allowed to take place very rapidly in air, the temperature of the surrounding air will

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Remains constant

Increase

Decrease

Either increase or decrease

One Joule is equivalent to

According to Hess's law, the enthalpy change for a reaction

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Depends on path

Independent of the path

The sum of ∆E and ∆H

None of these

Calorie is equivalent to

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0.4184J

4.184J

418.4J

40.18J

Enthalpy of neutralization of strong acids and strong bases have same values because

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Neutralization leads to the formation of salt and water

Acids always give rise to H+ and bases always furnish OH-

Strong acids and bases are ionic substances

The net change involves the combination of H and OH ions to form water

The enthalpies of all elements in their standard states are

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Unity

always +ve

always -ve

zero

One Joule is equivalent to

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4.184 cal.

0.4184cal.

1/2 cal.

1/4.184 cal

The values of ∆H for the process l(g)+e-1--------->
I-1(g) is:

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>0

<0

0

None

A system absorbs 100 kJ heat and performs 50 kJ work on the surroundings. The increase in internal energy of the system is

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50kJ

100 kJ

150kJ

5000 kJ

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