Chemistry · Physics

Thermodynamics and Chemical Kinetics

1,132 Questions

Thermodynamics and chemical kinetics are crucial branches of physical chemistry. This area focuses on energy transformations, reaction rates, equilibrium constants, and catalysts. These topics carry significant weight in competitive science and engineering examinations.

Enthalpy and energyChemical equilibriumEntropy conceptsReaction kinetics and catalysts

Thermodynamics and Chemical Kinetics Questions

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Which are correct representation at equilibrium?

  1. $\displaystyle p=\frac { eRT }{ N } $
  2. $\displaystyle K={ e }^{ { { -\Delta G }^{ o } }/{ RT } }$
  3. $\displaystyle \frac { { K } _{ 1 } }{ { K } _{ 2 } } ={ e }^{ { { -E } _{ a } }/{ RT } }$
  4. $\displaystyle \frac { P }{ { P }^{ o } } ={ e }^{ { -\Delta H }/{ RT } }$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $\displaystyle { K } _{ c }$ is:

  1. $\displaystyle { \Delta G }^{ o }=RTIn{ K } _{ c }$
  2. $\displaystyle -{ \Delta G }^{ o }=RTIn{ K } _{ c }$
  3. $\displaystyle { \Delta G }=RTIn{ K } _{ c }$
  4. $\displaystyle -{ \Delta G }=RTIn{ K } _{ c }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle \because \quad \Delta G={ \Delta G }^{ o }+RTInQ$, where Q is reaction quotient at equilibrium, $\displaystyle \Delta G=0$ and $\displaystyle Q={ K } _{ c }$
$\displaystyle \therefore \quad -\Delta G=RTIn{ K } _{ c }$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Which are true for the reaction: $A _2\rightleftharpoons 2C+D$?

  1. If $\Delta H=0; K _p$ increases with temperature and dissociation temperature.
  2. If $\Delta H=+ve; K _p$ increases with temperature and dissociation of $A _2$ increases.
  3. If $\Delta H=-ve; K _p$ increases with temperature and dissociation of $A _2$ decreases.
  4. $K _p=4\alpha^3\left [\frac {P}{1+2\alpha}\right ]^2$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Initial At equilibrium
$\underset {\underset {1-\alpha}{1}}{A _2}\rightleftharpoons \underset {\underset {2\alpha}{0}}{2C}+\underset {\underset {\alpha}{0}}{D}$
$K _p=(2\alpha)^2\alpha \times \left [\frac {P}{\Delta n}\right ]^2=\frac {4\alpha^3P^2}{(1+2\alpha)^2}$
Also, as we know
2.303 log $\frac {K _2}{K _1}=\frac {\Delta H}{R}\left [\frac {T _2-T _1}{T _1T _2}\right ]$ for effect temperature on K.
If $\Delta H=+ve; K _p$ increases with temperature and dissociation of $A _2$ increases.
If $\Delta H=-ve; K _p$ decreases with temperature and dissociation of $A _2$ decreases.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $K$ is 

  1. $-\Delta G=RT\:\ln\:K$
  2. $\Delta G^{o}=RT\:\ln\:K$
  3. $\Delta G=-RT\:\ln\:K$
  4. $-\Delta G^{o}=RT\:\ln\:K$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant K is $-\Delta G=RT:ln:K$ or $\Delta G=-RT:ln:K$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

For the reaction at $298 K$


$A (g) + B (g)\rightleftharpoons C (g) + D (g)$

$\Delta H^o = 29.8 kcal ; \Delta S^o = 0.1 kcal/K$

Calculate $\Delta G^o$ and $K$.

  1. $\Delta G^o = 0 ; K = 1$
  2. $\Delta G^o = 1 ; K = e$
  3. $\Delta G^o = 2 ; K = e^2$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As we know,


$\Delta G^o = \Delta H^o - T\Delta S^o$ 

         $= 29.8 - ( 298\times0.1 )$

         $= 29.8-29.8=0$

Therefore, $\Delta G^o = 0$

The relation between $\Delta G^0 $ and $K$

$\Delta G^0$ = $ - RT lnK$

$K = 1 $

So, the correct option is $A$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

When $\displaystyle \Delta G$ is zero :

  1. reaction moves in forward direction

  2. reaction moves in backward direction

  3. system is at equilibrium

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When $\displaystyle \Delta G$ is zero, system is at equilibrium.
Positive free energy change corresponds to non spontaneous reaction.
Negative free energy change corresponds to spontaneous reaction.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Which is not correct relationship between $\Delta G^{ \ominus }$ and equilibrium constant $K _P$

  1. $K _P = -RT log \Delta G^{ \ominus }$
  2. $K _P = [e/RT]^{ \Delta G^{ \ominus } }$
  3. $K _P = -\frac { \Delta G^{ \ominus } }{ RT }$
  4. $K _P = e^{ -\Delta G^{ \ominus }/RT }$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$\Delta G = \Delta G^{ \ominus } + RT log K$
and at equilibrium,
$\Delta G = 0$ so
$\Delta G^{ \ominus } = - RT log K$
$\Delta G^{ \ominus } = -RTln K _P$
$K _P = e^{ -\Delta G^{ \ominus }/RT }$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

The correct relation between equilibrium constant $(K)$, standard free  energy $(\Delta {G}^{o})$ and temperature $(T)$ is:

  1. $\Delta {G}^{o}=RT\ln {K}$
  2. $K={ e }^{ \Delta { G }^{ o }/2.303 RT }\quad $
  3. $\Delta { G }^{ o }=-RT\log{K}$
  4. $K={ 10 }^{ -\Delta { G }^{ o }/2.303 RT }\quad $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Consider a reaction, $A+B\rightleftharpoons C+D$

$\Delta G={ \Delta G }^{ 0 }+RTlnQ$
for equilibrium, $\Delta G=0$
$\therefore \quad 0={ \Delta G }^{ 0 }+RTlnK$
$\therefore \quad { \Delta G }^{ 0 }=-RTlnK$
$\therefore \quad { \Delta G }^{ 0 }=-2.303RTlogK$
i.e. $K={ 10 }^{ { -\Delta G }^{ 0 }/2.303RT }$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

If we know $\displaystyle { \Delta G }^{ \circ  }$ of a reaction, which of the following can be defined ?
I. Cell potential, $\displaystyle { E }^{ \circ  }$
II. Activation energy, $\displaystyle { E } _{ a }$
III. Equilibrium constant, $\displaystyle { K } _{ eq }$

  1. I and II only

  2. I and III only

  3. III only

  4. I, II, III

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If we know $\displaystyle \Delta G^o $  of a reaction, the following can be defined. 
I. Cell potential, $\displaystyle E^o $
III. Equilibrium constant, Keq
$\displaystyle \Delta G^o = -nF E^o = - RTlnK $

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

By which of the following relations, the equilibrium constant varies with temperature?

  1. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =\cfrac { \Delta { H }^{ o } }{ R } \int _{ { T } _{ 1 } }^{ { T } _{ 2 } }{ d\left( \cfrac { 1 }{ T } \right) } $
  2. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =-\cfrac { \Delta { H }^{ o } }{ R } \int _{ { 1/T } _{ 1 } }^{ { 1/T } _{ 2 } }{ d\left( \cfrac { 1 }{ { T }^{ 2 } } \right) } $
  3. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =-\cfrac { \Delta { H }^{ o } }{ R } \int _{ { T } _{ 1 } }^{ { T } _{ 2 } }{ d\left( \cfrac { 1 }{ T } \right) } $
  4. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =-\cfrac { \Delta { H }^{ o } }{ R } \int _{ { 1/T } _{ 2 } }^{ { 1/T } _{ 1 } }{ d\left( \cfrac { 1 }{ { T }^{ } } \right) } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\textbf{Explanation:}$

  • We can use $\mathit{Gibbs-Helmholtz}$ to get the temperature dependence of $K$                                                                                            
$\mathbf{\left ( \frac{\partial \left [ \Delta _{r}G^{o} \right ]}{\partial T} \right )}$   $\mathbf{=\frac{-\Delta _{r}H^{o}}{T^{2}}}$     $\mathbf{\rightarrow \left ( 1 \right )}$

  • At equilibrium, we can equate $\Delta _{r}G^{o}$ to $-RTlnK$ so we get

  $\mathbf{\left ( \frac{\partial \left [ lnK \right ]}{\partial T} \right )= \frac{\Delta _{r}H^{o}}{RT^{2}}}$    $\mathbf{\rightarrow (2)}$

  • We see that whether  K  increases or decreases with temperature is linked to whether the reaction enthalpy is positive or negative. If the temperature is changed little enough that  $\Delta _rH^{o}$  can be considered constant, we can translate a  $K$  value at one temperature into another by integrating the above expression, we get a similar derivation as with melting point depression:

$\mathbf{ln\frac{K\left ( T _{2} \right )}{K\left ( T _{1} \right )}=\frac{-\Delta _{r}H^{o}}{R}\left ( \frac{1}{T _{2}}-\frac{1}{T _{1}} \right )}$$\mathbf{\rightarrow \left ( 3 \right )}$

  • If we integrate and differentiate the left side of the equation then we get and solve the right side we get

$\mathbf{lnK _{2}-lnK _{1}=\frac{-\Delta H^{0}}{R}\int _{T _{1}}^{T _{2}}\mathbf{\mathit{d}}\left ( \frac{1}{T} \right )}$$\mathbf{\rightarrow (4)}$

Hence from equation $4$ we can say that option $C$ is correct.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

In dynamic equilibrium condition, the reaction on both the sides occurs at the same rate and the mass on both sides of the equilibrium does not undergo any change. This condition can be achieved only when the value of $\Delta$G is :

  1. -1

  2. +1

  3. +2

  4. 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The Gibb's free energy change ($\Delta G$) is:
$\Delta G<0 \longrightarrow$ Spontaneous process
$\Delta G>0 \longrightarrow$  Non-spontaneous process
$\Delta G=0 \longrightarrow$ Equilibrium process
Therefore, the condition of dynamic equilibrium can be achieved only when $\Delta G=0$.