Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

(b) According to Gibbs-Helmholtz equation, Gibbs energy ($∆$G) = $∆$H - T$∆$S

where, $∆$H = Enthalpy change

$∆$S = Entropy change

T = Temperature

For a reaction to be spontaneous $∆$G <0.

$\therefore$Gibbs -Helmholtz equation becomes.

$∆$G = $∆$H -T$∆$S<0

or, $∆$H < T$∆$S
T > $∆$H / $∆$S

=35.5 KJmol^-1/ 83.6 JK^-1mol^-1

=35.5 x 1000J mol^-1 / 83.6JK^-1mol^-1

=425 K

T>425K

(c) Key concept According to first law of thermodynamics,

$∆$U = q + w

where, $∆$U =internal energy

q = heat absorbed or evolved, w = work done.

Also, work done against constant external pressure (irreversible process).

W = -P_ext $∆$V.

Work done in irreversible process,

w = -P_ext $∆$V = - p_ext (V₂ - V₁)

= -2.5 atm (4.5 L - 2.5 L)

= - 5 L atm = - 5 x 101.3 J

= - 505 J

Since, the system is well insulated, q =0

$∆$U = w = - 505 J

Hence, change in internal energy, $∆$U of the gas is - 505 J.

(b) Entropy change is given as,

     $∆$S = nC_pln(T_f/T_i) + nRln(p_i/p_f)           ....(i)

For isothermal process, T_i = T_f

.^..  nC_pln(T_f/T_i) =0 [ln1 = 0]

From Eq (i)  $∆$S = nRln(p_i/p_f)

Given, C(s) + O₂(g) $\to$CO₂(g);

$∆$_fH = -393.5 kJ mol^-1

._.. Heat released on formation of 44 g or 1 mole

               CO₂ = -395.5 kJ mol

.^..  Heat released on the formation of 35.2 g of CO₂

        = -393.5 kJ mol^-1/44 g  x 35.2 g = -315 kJ mol^-1

The change in Gibbs free energy is given by 

$∆\mathrm{G} = ∆\mathrm{H} - \mathrm{T}∆\mathrm{S}$

where, $∆\mathrm{H}$ = enthalpy of the reaction 

$∆$S = entropy of the reaction

Thus, in order to determine $∆$ the value of $∆$ must be known. The value of $∆$ can be calculated by the equation

$∆$H = $∆$U + $∆$n_gRT

Where $∆$U = Change  in internal energy.

$∆$n_g = (number of moles of gaseous product) - (number of moles of gaseous rectant)

= 2-0 = 2

R = gas constant = 2cal

But, $∆$H = $∆$U + $∆$n_gRT

$∆$U =2.1Kcal = 2.1 x 10³ cal ( 1kcal = 10³cal)

$∆$H = (2.1x10³)+(2x2x300) = 3300 cal

Hence, $∆$G = $∆$H -T$∆$S

$∆$G = 3300 - (300x20)

Among the given reactions only in the case of 

 C(graphite) + 1/2 O₂(g) $\to$CO(g)

entropy increases because randomness (disorder) increases. Thus, standard entropy change ($∆$S$°$) is positive.

Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, ie, 

$∆$H$°$=-ve

We know that

   $$\begin{gathered} ∆\mathrm{G}°=∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}° \\ ∆\mathrm{G}° = -\mathrm{ve} -\mathrm{T}(+\mathrm{ve}) \end{gathered}$$

Thus, as the temperature increases, the value of $∆$G$°$ decreases.

Molar entropy change for the melting of ice, 

$∆{\mathrm{S}}_{\mathrm{vap}.}=∆{\mathrm{H}}_{\mathrm{vap}. }/ \mathrm{T}$

= (1.435Kcal/mol) / (0+273)K

= 5.26 x 10^-3 Kcal/molK

= 5.26 cal/mol K

H₂O(l) $\stackrel{100°\mathrm{C}}{\to }$H₂O(g)

${∆}_{\mathrm{vap}}\mathrm{H}° = {∆}_{\mathrm{vap}}\mathrm{E}° + ∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT}$

$∆$n_g = n_p - n_r = 1-0=1

.^.. 40.66 kJ mol^-1 = $∆$_vapE$°$ + 1x8.314x10^-3x373

$∆$_vapE$°$ = 40.66 kJ mol^-1 -3.1 kJ mol^-1

             = +37.56 kJ mol^-1

$∆\mathrm{G}° = ∆\mathrm{H}° - \mathrm{T}∆\mathrm{S}°$

Given, $∆{\mathrm{H}}_{\mathrm{vap}. }$ = 30 KJmol^-1

$∆\mathrm{G}°$ = 0 at equilibrium,

$∆{\mathrm{S}}_{\mathrm{vap}.}=∆{\mathrm{H}}_{\mathrm{vap}. }/ \mathrm{T}$

= (30x10³Jmol^-1 ) / 300K

= 100 Jmol^-1k^-1

4H(g) $\to$2H₂(g); $∆$H = -869.6 kJ

2H₂(g) $\to$4H(g); $∆$H = 869.6 kJ

H₂(g) $\to$2H(g); $∆$H = 869.6/2 = 434.8 kJ