Ncert Solutions Chemistry Class 11th
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1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.
1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.
This is a Short Answer Type Questions as classified in NCERT Exemplar
We can conclude from the figure that this change is a reversible change.
Now,
W= -2.303nRT log
But, p1V1 = p2V2 = = = =2
W= -2.303nRT log
= -2.303 *8.314*1*298*log2
= -2.303 *8.314*298*0.3010J
= -1717.46J
Upvote 
Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide spontaneity of a reaction from given diagram. Explain
Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide spontaneity of a reaction from given diagram. Explain
This is a Short Answer Type Questions as classified in NCERT Exemplar
No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.
Upvote Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof.
(b) 12H2(g) + 12Cl2(g) HCl(g) ?rH? = –92.32 kJ/ mol
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof.
(b) 12H2(g) + 12Cl2(g) HCl(g) ?rH? = –92.32 kJ/ mol
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
This is a Short Answer Type Questions as classified in NCERT Exemplar
Throwing a stone from ground to roof
b) the reaction involved is a process where the energy decreases after the reaction. It can be represented as:In process b), potential energy/enthalpy change is a contributing factor to the spont
Upvote How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps ?
How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps ?
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When a process can be reversed by bringing an extremely small change in it, we call it a reversible process. The pressure-volume graph can be used to calculate the work done. The pressure is not constant, and changes in infinitesim
Upvote What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, pext in a single step as shown in Fig. 6.2. Explain graphically
What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, pext in a single step as shown in Fig. 6.2. Explain graphically
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Assumption: Cylinder is filled with one mole gas, and the piston is frictionless. Let the pressure of gas inside be p and the volume of gas be V_ {I}.
Piston is moved towards the inside to make the external pressure (P_ {ext}) equal
Upvote The enthalpy of reaction for the reaction : 2H2(g) + O2(g) → 2H2O(l) is ∆rHᶱ = – 572 kJ/ mol. What will be standard enthalpy of formation of H2O (l ) ?
The enthalpy of reaction for the reaction : 2H2(g) + O2(g) → 2H2O(l) is ∆rHᶱ = – 572 kJ/ mol. What will be standard enthalpy of formation of H2O (l ) ?
Chemistry NCERT Exemplar Solutions Class 11th Chapter six
Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given eq
Upvote The enthalpy of vapourisation of CCl4 is 30.5 kJ/ mol. Calculate the heat required for the vapourisation of 284 g of CCl4 at constant pressure. (Molar mass of CCl4 = 154 g/mol).
The enthalpy of vapourisation of CCl4 is 30.5 kJ/ mol. Calculate the heat required for the vapourisation of 284 g of CCl4 at constant pressure. (Molar mass of CCl4 = 154 g/mol).
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As per the information provided in the question, for one mole of CCl4 (154 g), the heat of vaporisation required is 30.5 kJ/mol .
Hence for the vaporisation of 284 g of CCl4, we require:
= 56.2 kJ
Upvote The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.
H2(g) + Br2(g) → 2HBr(g)
Given that Bond energy of H2, Br2 and HBr is 435 kJ/ mol, 192 kJ/ mol and 368 kJ/ mol respectively
The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.
H2(g) + Br2(g) → 2HBr(g)
Given that Bond energy of H2, Br2 and HBr is 435 kJ/ mol, 192 kJ/ mol and 368 kJ/ mol respectively
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ΔrH? =ΣB.E (reactant)-ΣB.E (product)
=B.E.H2 + BEBR2 -2 ×B.E>HBr
= 435+192-2 (×368)
= 109KJmol-1
Upvote Using s, p and d notations, describe the orbitals with following quantum numbers:
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Using s, p and d notations, describe the orbitals with following quantum numbers:
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital
Upvote (i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers ml and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f.
(i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers ml and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f.
(i) For n = 3; l = 0, 1 and 2.
For l = 0 ; ml = 0
For l = 1; ml = +1, 0, -1
For l = 2 ; ml = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; ml can have any of the values -2, -1, 0, + 1, +2.
(iii) For 1p orbital, n and l are both e
Upvote