Ncert Solutions Chemistry Class 11th

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New answer posted

5 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

In order to calculate the lattice enthalpy of NaBr,

(i) Na (s) →Na (g) ; ΔsubH? =108.4 kJ/mol

(ii) Na→Na+ + e-   ΔiH? = 496kJ/ mol

(iii) 1 2 Br→ Br, 1 2 Δdiss H? = 96kJ/ mol

(iv) Br+e-Br-  ΔegH? = - 325 kJmol-1

? fH? =? subH? + Δdiss H + Δi H? + Δi H? + Δeg H? +? lattice H?

= -360.1 -108.4-96-496+325 = -735.5KJ/ mol

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

The reaction presented in the question is

CH4 (g)→C (g) + 4H (g)

Now, ΔaH = 1665  kJ/mol

The mean bond enthalpy of the C-H bond should be used here. For the atomisation of 4 moles of C-H bonds, the value is 1665 kJ/mol. So, per mole energy = 1665/4 = 416.2 kJ / mol

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

According to Hess's law, ΔrH = ΔrH1+ ΔrH2rH3

This is so because during the reaction A→ B, B's formation undergoes various intermediate reactions, with the overall value of the enthalpy being ΔrH.

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5 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

ΔH for formation is given. For the reverse reaction, ΔHchanges sign as the reverse of exothermic reaction will be endothermic. So, ΔH for decomposition is - (-91.8)=91.8 for one mole. But here, two moles are decomposing,

ΔH=2*91.8=183.6KJ

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Standard molar enthalpy of formation,  rH1- is just a special case of fH2-, where

one mole of a compound is formed from its constituent elements. In the above equation, enthalpy of formation and enthalpy of reaction is not the same.

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Due to weak force of attraction between molecules, acetone requires less heat to vaporise. Hence, water has higher enthalpy of vaporization.

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5 months ago

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Enthalpy of a reaction is the energy change per mole for the process.

18 g of H2O = 1 mole ΔHvap = 40.79 kJ/ mol

 Enthalpy change for vapourising 2 moles of H2O = 2 x 40.79 = 81.58 kJ ΔH°vap = 40.79 kJ mol-1.

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

(i) Reversible work is represented by the combined areas

(ii) Work against constant pressure, Pis represented by the area

          

Work (i) > Work (ii). 

The Approach While Dealing With the Concept of Thermodynamics

Since the concept of Thermodynamics and the terminologies of Chemistry are a bit new to the students, they should first learn the names of different chemical components and how to write the chemical equations properly through NCERT Exemplars and Solutions books.

While writing the chemical equations, they might make some mistakes. To avoid making errors, they shoul

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5 months ago

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

We know,

ΔStotal = ΔSsys + ΔSsurr

When a system is in thermal equilibrium with its surroundings, the surroundings' temperature is the same as the system's. Furthermore, a rise in the enthalpy of the surroundings equals a decrease in the system's enthalpy. As a result of the entropy shift in the environment,

ΔSsurr = ΔHsurr/T = -ΔHsys/T

ΔStotal = ΔSsys = (-ΔHsys/T)

Rearranging the above equation:

ΔStotal = TΔSsys - ΔHsys

For spontaneous process,

ΔStotal > 0, so 

TΔSsys - ΔHsys> 0

⇒ ( - ΔHsys - TΔSsys) > 0

The above equation can be written as

- ΔG > 0

ΔG

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New answer posted

5 months ago

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

A contrast is drawn in thermodynamics between extensive and intense qualities. An extensive property is one whose value is proportional to the amount or size of matter in the system. Extensive properties include mass, volume, internal energy, enthalpy, and heat capacity, to name a few.

Properties that are independent of the amount or size of matter present are known.

As though they were intensive properties Temperature, density, and pressure, for example, are intense properties. A molar property? m, is the value of an extensive property of the system for 1 mol of the sub

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