Chemistry MCQs for NEET — Practice Questions with Answers

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

Key Idea For spontaneous process $∆$G<0

$∆$G=$∆$H-T$∆$S

Given, $∆$H = 170 kJ = 170 x 10³ J

           $∆$S = 170 JK^-1

           T=?

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

 0<170 x10³ - Tx170

          T>1000

.^.. T=1110 K

Key Idea: State function is the property of the system whose value depends only on the initial and final state of the system and is independent of the path.

._.. Internal energy ($∆$E) = q + W

It is a state function because it is independent of the path. It is an extensive property.

._.. Gibbs energy (G) = H-TS

It is also a state function because it is independent of the path. It is also an extensive property. Heat (q) and Work (W) are not state functions being path dependent.

Key idea:  ΔH_reaction =  Σ Bond energy of reactant -  Σ Bond energy of the product

Here, ΔH_H-H = 434 kJ mol^-1

        ΔH_Cl-Cl = 242 kJ mol^-1

        ΔH_H-Cl = 431 kJ mol^-1

        1/2H₂=1/2Cl₂ $\to$ HCl

ΔH_reaction = 1/2ΔH_H-H + 1/2ΔHCl-Cl - ΔH_H-Cl

               = (1/2)x434 = (1/2)x242-431

               =  217+121-431

               = -93 kJ mol^-1