Thermodynamics

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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 different stage it would vary.

ΔH = 0

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ΔrG° = -RT ln Kp

= -RT ln (0.98)

Since In (0.98) is negative

.'. ΔrG° is positive the reaction is non spontaneous

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Yes, when the system and the surroundings are in thermal equilibrium, their temperatures are the same.

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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 = q_rev/ T

Here? S = change in entropy  ^

q_rcv = heat of reversible reaction '

T = temperature

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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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In order to calculate the lattice enthalpy of NaBr,

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

(ii) Na→Na⁺ + e^-   Δ_iH? = 496kJ/ mol

(iii) $\mathrm{}\frac{1}{2}$ Br→ Br, $\mathrm{}\frac{1}{2}$ Δ_diss H? = 96kJ/ mol

(iv) Br+e^-Br^-  Δ_egH? = - 325 kJmol^-1

? _fH? =? _subH? + Δ_diss H + Δ_i H? + Δ_i H? + Δ_eg H? +? _lattice H?

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

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The reaction presented in the question is

CH₄ (g)→C (g) + 4H (g)

Now, Δ_aH = 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

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According to Hess's law, Δ_rH = Δ_rH₁+ Δ_rH₂+Δ_rH₃

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

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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

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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 same.