Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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

(a) Enthalpy of formation: The amount of heat evolved or absorbed during the formation of 1 mole of a compound from its constituent elements is known as the heat of formation. So, the correct answer is:

${\mathrm{H}}_2\left(\mathrm{g}\right) + \frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) \stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right), ∆\mathrm{H}=-{\mathrm{x}}_2{\mathrm{kJmol}}^{-1}$

$$\begin{gathered} \mathrm{Given} {\mathrm{H}}_2 \left(\mathrm{g}\right) \to 2\mathrm{H} \left(\mathrm{g}\right) = +430 .....\left(1\right) \\ {\mathrm{Cl}}_2 \left(\mathrm{g}\right) \to 2\mathrm{Cl} \left(\mathrm{g}\right) +240 ......\left(2\right) \\ \frac{1}{2}{\mathrm{H}}_2 \left(\mathrm{g}\right) + \frac{1}{2}{\mathrm{Cl}}_2 \left(\mathrm{g}\right) \to \mathrm{HCl} \left(\mathrm{g}\right) ∆{\mathrm{H}}_{\mathrm{f}}=-90 .........\left(3\right) \\ \mathrm{Muliplying} \left(1\right) \& \left(2\right) \mathrm{by} 1/2 \\ \left(4\right) \frac{1}{2}{\mathrm{H}}_2 \left(\mathrm{g}\right) \to \mathrm{H} \left(\mathrm{g}\right) = \frac{430}{2}=215 \\ \left(5\right) \frac{1}{2}{\mathrm{Cl}}_2 \left(\mathrm{g}\right) \to \mathrm{Cl} \left(\mathrm{g}\right) = \frac{240}{2}=120 \\ \mathrm{Adding} \left(4\right) \& \left(5\right) \\ = \frac{1}{2}{\mathrm{H}}_2 \left(\mathrm{g}\right) + \frac{1}{2} {\mathrm{Cl}}_2 \left(\mathrm{g}\right) \to \mathrm{H} \left(\mathrm{g}\right)+\mathrm{Cl} \left(\mathrm{g}\right) \\ = 215+120 = 335 ........\left(6\right) \\ \mathrm{Substracting} \mathrm{eq}. \left(3\right) \mathrm{from} \mathrm{eq}. \left(6\right) \mathrm{we} \mathrm{get}, \\ \frac{1}{2}{\mathrm{H}}_2 \left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Cl}}_2 \left(\mathrm{g}\right) \to \mathrm{H} \left(\mathrm{g}\right)+\mathrm{Cl} \left(\mathrm{g}\right) 335 \\ \frac{1}{2}{\mathrm{H}}_2 \left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Cl}}_2 \left(\mathrm{g}\right) \to \mathrm{HCl} \left(\mathrm{g}\right) \left(-\right)\left(-90\right) \\ \overline{ \mathrm{HCl} \left(\mathrm{g}\right) \to \mathrm{H} \left(\mathrm{g}\right) + \mathrm{Cl} \left(\mathrm{g}\right) = 335+90 } \\ = 425 \end{gathered}$$

If the Gibbs free energy for a system ( $∆$G_system) is equal to zero, then the system is present in equilibrium at a constant temperature and pressure.

At equilibrium Gibbs free energy change (ΔG°) is equal to zero. The following thermodynamic relation is used to show the relation of ΔG° with enthalpy change (ΔH°) and entropy change(ΔS°)

ΔG° = ΔH°-TΔS

0 = 30 x 10³ (J mol^-1) - T x 105 (J K^-1 mol^-1)

T = 3X10³/105 K= 285.71 K

Δng=0, ΔH° = ΔU° = – 185 KJ

For 3 mole, ΔU° = 3 x (– 185) = – 555 KJ

0.42 = a(10)³⇒a = 0.42 × 10³

S_m = ${\int }_0^{10}\frac{{\mathrm{C}}_{\mathrm{pm}}}{\mathrm{T}}$ dT = $\underset{0}{\overset{10}{\int }}{\mathrm{aT}}^2\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{\mathrm{a}}{3}[{10}^3\mathrm{\hspace{0.17em}}-\mathrm{\hspace{0.17em}}0]\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{0.42}{3}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}0.14\mathrm{J}/\mathrm{K}\hspace{0.17em}-\hspace{0.17em}\mathrm{mol}$  

ΔS = nR ln $\left(\frac{{\displaystyle V_f}}{{\displaystyle V_i}}\right)=R\ln \left(\frac{{\displaystyle P_i}}{{\displaystyle P_f}}\right)$ = R ln $\left(\frac{{\displaystyle 300\mathrm{R}}}{{\displaystyle 1\mathrm{L}\times 1\hspace{0.17em}\mathrm{atm}}}\right)$ = Rln (24.6)