Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

(a) For reaction,

C(s) + 2H₂(g) $\to$CH₄(g) , $∆\mathrm{H}°$=?

 $∆\mathrm{H}°$=-[$∆{\mathrm{H}}_{\mathrm{c}}^{°}$ of CH₄) -($∆{\mathrm{H}}_{\mathrm{c}}^{°}$of C +2x $∆{\mathrm{H}}_{\mathrm{c}}^{°}$ of H₂)]

              C+ O₂ $\to$CO₂,    $∆\mathrm{H}°$ = -94 kcal   ...(i)

           2H₂ + O₂ $\to$2H₂O,  $∆\mathrm{H}°$= -68 x 2 kcal         ...(ii)

     CH₄ + 2O₂ $\to$CO₂ + 2H₂O,  $∆\mathrm{H}°$=-213 kcal    .....(iii)

On adding eqs. (i) and(ii) and then subtracting eq (iii)

              = -[(-213)-(-94+2x-68)] kcal/mol

              = -[-213 + 230] = -17 kcal/mol

(a) q=$∆\mathrm{U} + \mathrm{W}$, its work done by the system.

    50=$∆$U + 30

 .^..   $∆$U = 20 cal

(d) In a cyclic process, $∆\mathrm{U}=0$

(d) The gaseous phase have more entropy and thus, $∆$S is +ve in (a) and (b). Also decrease in pressure increase disorder and thus, $∆$S is +ve in (c). In (d) the disorder decreases in liquid state due to the decrease in temperature. Thus, $∆$S= -ve

(b) If work done by the system is positive, then q=$∆$U+W. However, new terminology has revealed that work done by the system is negative and work done on the system is positive. Thus, according to this, q=$∆$E-W.

(d) These all are facts.

(b) Gaseous product is formed; more positive is $∆\mathrm{n}$, more is entropy.

(b) $$\begin{gathered} ∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}; \mathrm{at} \mathrm{equilibrium}, \\ ∆\mathrm{G}=0, {.}^{.}.∆\mathrm{H}=\mathrm{T}∆\mathrm{S} \end{gathered}$$

       or $∆\mathrm{H}$ = 273 x (60.01-38.20) = 5954.13 J mol^-1

(b) The system returns to its original state, i.e, cyclic process.

(a) TV ^r-1 =K

      T₁${\mathrm{V}}_1^{\mathrm{r}-1}$= T₂${\mathrm{V}}_2^{\mathrm{r}-1}$

      T₁${\mathrm{V}}_1^{2/3}$= T₂${\mathrm{V}}_2^{2/3}$                       (r=5/3 for monoatomic)

         T₁/T₂ = (V₂/V₃)^2/3   = (1/2)^2/3 

           .^.. T₂ = T/(2)^2/3