NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | CHEMISTRY IMPORTANT QUESTIONS PART-1 | CHAPTER -6 | THERMODYNAMICS | EDUGROWN |

In This Post we are  providing Chapter-6 THERMODYNAMICS NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON THERMODYNAMICS

1.With the help of first law of thermodynamics and H = U + pv, prove = q_p

Ans.The enthalpy is defined as

H = U + pv

For a change in the stales of system,

= 

= 

= …………(i)

The first law of thermodynamics states that –

= …………………….(ii)

From (i) and (ii),

= 

When the pressure is constant,

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2.Why is the difference between and not significant for solids or liquids?

Ans. The difference between and is not usually significant for systems consisting of only solids and / or liquids because they do not suffer any significant volume changes upon heating.

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3.What is an extensive and intensive property?

Ans. Extensive property is a property whose value depends on the quantity or size of matter present in the system.

Intensive property is a property which do not depend upon the quantity or size of matter present.

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4.Calculate the heat of combustion of ethylene (gas) to from CO₂ (gas) and H₂O (gas) at 298k and 1 atmospheric pressure. The heats of formation of CO_2, H₂O and C₂H₄ are – 393.7, – 241.8, + 52.3 kJ per mole respectively.

Ans. C₂H₄ (g) + 30₂(g) 2CO₂(g) + 2H₂O (g)

reactants

= [2 x (CO₂) + 2 x ] – 

= 2 x[(-393.7)m+2x (-241.8)] – [(523.0) + 0)]

= [-787.4 – 483.6 ] -53.3

= – 1323.3 kJ.

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5.Give two examples of reactions which are driven by enthalpy change.

Ans. Examples of reactions driven by enthalpy change:

The process which is highly exothermic, i.e. enthalpy change is negative and has large value but entropy change is negative is said to be driven by enthalpy change, eg.

(i)

(ii)

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6.Will the heat released in the following two reactions be equal? Give reasons in support of your answer.

(i)H₂ (g) + 

Ans. No, the heats released in the two reactions are not equal. The heat released in any reaction depends upon the reactants, products and their physical states. Here in reaction (i), the water produced is in the gaseous state whereas in reaction (ii) liquid is formed. As we know, that when water vapors condensed to from water, heat equal to the latent heat of vaporization is released. Thus, more heat is released in reaction (ii).

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7.What is the relation between the enthalpy of reaction and bond enthalpy?

Ans .A chemical reaction involves the breaking of bonds in reactants and formation of new bonds in products. The heat of reaction (enthalpy change) depends on the values of the heat needed to break the bond formation .Thus

(Heat of reaction = (Heat needed to break the bonds in reactants – Heat liberated to from bonds in products).

= Bond energy in (to break the bonds) – Bond energy out (to form the bonds)

= Bond energy of reactants – Bond energy of products.

8.The are given + 61.17kJ mol^-1 and + 132 Jk^-1mol^-1 respectively. Above what temperature will the reaction be spontaneous?

Ans. The reaction

2Ag₂O (s) 

Will be spontaneous when is negative.

Since 

Shows that would be –ve when,

Or T > 

The process will be spontaneous above a temperature of .

9.Give the relationship between for gases.

Ans.For gases the volume change is appreciable.

let V_A be the total volume of gaseous reactants, and

V_B be the total volume of gaseous product.

n_A be the number of moles of the reactant and

n_B be the number of moles of the product,

Then at constant pressure and temperature,

p V_A = n_A RT

p V_B = n_B RT

or p V_B – pV_A = (n_B – n_A) RT

or p 

where and is equal to the difference between the number of moles of gaseous products and gaseous reactants.

Substituting the value of p awe get.

(heat change under constant pressure)

(heat change under constant volume)

for gaseous system.

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10.It has been found that 221.4J is needed to heat 30g of ethanol from 15⁰C to 18⁰C. calculate (a) specific heat capacity, and (b) molar heat capacity of ethanol.

Ans.(a) Specific heat capacity

C = 

= 

Since 1⁰C is equal to 1k, the specific heat capacity of ethanol = 2.46Jg^-1 0c^-1.

(b) Molar heat capacity, Cm = specific heat x molar mass.

Therefore, Cm (ethanol) = 2.46 x 46

= 113.2 Jmol^-1 0c^-1

The molar heat capacity of ethanol is 113.2 J mol^-1 0c^-1.

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