Ncert Solutions Chemistry Class 11th

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No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.

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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 spontaneity.

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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 infinitesimal amounts as compression happens from initial volume V_i to the final volume V_f. The below graph depicts the work done with the shaded area.

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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 to p. Now, let us assume that this change takes place in a single step, hence, V is the final volume. The work done by the piston is depicted in the graph shown below by shading the area.

P_extΔV= AV₁   (V₁-V₂)

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 equation is – 572 kJ mol^–1

Now the enthalpy of formation for H₂O will be half the enthalpy of the value in the given equation. So now we can calculate that

? _fH? = $\frac{-572\mathrm{}\mathrm{K}\mathrm{J}}{2}$ = -286KJ/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:

$\frac{284}{154}*30.5$  = 56.2 kJ

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Δ_rH? =ΣB.E _(reactant)-ΣB.E _(product)

 =B.E._H2 + BE_BR2 -2 *B.E>_HBr

= 435+192-2 (*368)

= 109KJmol^-1

(a) 1s orbital

(b) 4f orbital

(c) 3p orbital

(d) 4d orbital

(i) For n = 3; l = 0, 1 and 2.

For l = 0 ; m_l = 0

For l = 1; m_l = +1, 0, -1

For l = 2 ; m_l = +2, +1,0, +1, + 2

(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0, + 1, +2.

(iii) For 1p orbital, n and l are both equal to 1. Since, l should always have a lower value than n. So, 1p ortial does not exist.

For 3f orbital, n=3 and l=3. For the same reason, the existence of 3f orbital is not possible. 1p and 3f orbitals are not possible

Number of electrons in:

(i) H₂⁺ = 1

(ii) H₂ = 2

(iii) O₂⁺ = 15