NEET Practice Questions (MCQs) with Answers & Solutions

$∆\mathrm{H}°=\frac{3}{2}∆{\mathrm{H}}_1^{°} + \frac{∆{\mathrm{H}}_2^{°}}{2}$

       = -713.7 - 177.5 = -891.2 KJ

 $∆\mathrm{H}°=2∆{\mathrm{H}}_1^{°} + 2\mathrm{x}∆{\mathrm{H}}_2^{°}-∆{\mathrm{H}}_3^{°}$

3. Target reaction C₂H₄ + H₂ $\to$C₂H₆

Given reaction

(i) 2C + 2H₂ $\to$C₂H₄    $∆{\mathrm{H}}_1$=x₁

(ii) 2C + 3H₂$\to$C₂H₆    $∆{\mathrm{H}}_2$= x₂

(ii)-(i) = x₂ - x₁

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{T}}=∆{\mathrm{S}}_{\mathrm{system}}+∆{\mathrm{S}}_{\mathrm{Surr}.} \\ = + \mathrm{ve} \end{gathered}$$

 $∆{\mathrm{S}}_{\mathrm{r}}$= 60-38.2=21.8 JK^-1 mol^-1

        .^.. $∆\mathrm{H}$=21.8 x 273 = 5951.4 J/mol

(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)