Ncert Solutions Chemistry Class 11th
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Given that ∆H= 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Given that ∆H= 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
This is a Short Answer Type Questions as classified in NCERT Exemplar
It is a spontaneous process. Although enthalpy change is zero, randomness or disorder (ΔS) increases and ΔS is positive. Therefore, in the equation, ΔG = ΔH – TΔS, the term TΔS will be negative. Hence ΔG will be negative.
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Use the following data to calculate ∆lattice Hᶱ for NaBr. ∆sub Hᶱ for sodium metal = 108.4 kJ/ mol
Ionization enthalpy of sodium = 496 kJ/mol
Electron gain enthalpy of bromine = – 325 kJ mol–1 Bond dissociation enthalpy of bromine = 192 kJ mol–1 ∆f Hᶱ for NaBr (s) = – 360.1 kJ/ mol
Use the following data to calculate ∆lattice Hᶱ for NaBr. ∆sub Hᶱ for sodium metal = 108.4 kJ/ mol
Ionization enthalpy of sodium = 496 kJ/mol
Electron gain enthalpy of bromine = – 325 kJ mol–1 Bond dissociation enthalpy of bromine = 192 kJ mol–1 ∆f Hᶱ for NaBr (s) = – 360.1 kJ/ mol
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) Br→ Br, Δdiss H? = 96kJ/ mol
(iv) Br+e-Br- ΔegH? = - 325 kJmol-1
? fH? =? subH?
Upvote The enthalpy of atomisation for the reaction CH4(g)→ C(g) + 4H(g) is 1665 kJ/ mol. What is the bond energy of C–H bond?
The enthalpy of atomisation for the reaction CH4(g)→ C(g) + 4H(g) is 1665 kJ/ mol. What is the bond energy of C–H bond?
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 m
Upvote Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆r H and ∆r H1, ∆rH2, ∆rH3 . represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆r H for overall reaction and ∆rH1 , ∆rH2 . etc. for intermediate reactions.
Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆r H and ∆r H1, ∆rH2, ∆rH3 . represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆r H for overall reaction and ∆rH1 , ∆rH2 . etc. for intermediate reactions.
This is a Short Answer Type Questions as classified in NCERT Exemplar
According to Hess’s law, ΔrH = ΔrH1+ ΔrH2+ΔrH3
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.
Upvote The value of ∆fHᶱ for NH3 is – 91.8 kJ/ mol. Calculate enthalpy change for the following reaction :
2NH3(g) → N2(g) + 3H2(g)
The value of ∆fHᶱ for NH3 is – 91.8 kJ/ mol. Calculate enthalpy change for the following reaction :
2NH3(g) → N2(g) + 3H2(g)
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=
Upvote Standard molar enthalpy of formation, ∆fHᶱ is just a special case of enthalpy of reaction, ∆r HV . Is the ∆r HV for the following reaction same as ∆f HV ? Give reason for your answer.
CaO(s) + CO2(g) → CaCO3(s); ∆fHᶱ = –178.3 kJ /mol
Standard molar enthalpy of formation, ∆fHᶱ is just a special case of enthalpy of reaction, ∆r HV . Is the ∆r HV for the following reaction same as ∆f HV ? Give reason for your answer.
CaO(s) + CO2(g) → CaCO3(s); ∆fHᶱ = –178.3 kJ /mol
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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
Upvote One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
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Due to weak force of attraction between molecules, acetone requires less heat to vaporise. Hence, water has higher enthalpy of vaporization.
Upvote 18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ/ mol. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?
18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ/ mol. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?
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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.
Upvote Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (Pi, Vi) to (Pf, Vf) . With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure Pf.
Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (Pi, Vi) to (Pf, Vf) . With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure Pf.
(i) Reversible work is represented by the combined areas
(ii) Work against constant pressure, Pf is represented by the area
Work (i) > Work (ii).
The Approach While Dealing With the Concept of Thermodynamics
Since the concept of TUpvote ΔG is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that ΔG is a measure of free energy. Find the unit of ΔG. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?
ΔG is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that ΔG is a measure of free energy. Find the unit of ΔG. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?
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
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