Thermodynamics
Get insights from 325 questions on Thermodynamics, answered by students, alumni, and experts. You may also ask and answer any question you like about Thermodynamics
- 325
Questions
- 0
Discussions
- 0
Active Users
- 0
Followers
Have a question about Thermodynamics? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Fig. 6.1. What will be the value of ?H for the cycle as a whole?
A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Fig. 6.1. What will be the value of ?H for the cycle as a whole?
This is a Short Answer Type Questions as classified in NCERT Exemplar
During cyclic process, change in internal energy is zero.
ΔU = 0
and no work is said to be done, as system returns to the initial state.
For a steady state cyclic process at any given stage enthalpy is one single value however, at dif
Upvote 
At 298 K. Kp for the reaction N2O4 (g) 2NO2 (g) is 0.98. Predict whether the reaction is spontaneous or not.
At 298 K. Kp for the reaction N2O4 (g) 2NO2 (g) is 0.98. Predict whether the reaction is spontaneous or not.
This is a Short Answer Type Questions as classified in NCERT Exemplar
ΔrG° = -RT ln Kp
= -RT ln (0.98)
Since In (0.98) is negative
.’. ΔrG° is positive the reaction is non spontaneous
Upvote Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
This is a Short Answer Type Questions as classified in NCERT Exemplar
Yes, when the system and the surroundings are in thermal equilibrium, their temperatures are the same.
Upvote Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
This is a Short Answer Type Questions as classified in NCERT Exemplar
Heat has a randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematical relation which relates these three parameters is? S = qrev/ T
Here? S = change in ent
Upvote 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.
Upvote 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
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
Upvote