Tag: entropy and spontaneity

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

The equilibrium constants of a reaction is $73$. Calculate standard free energy change.

  1. $-106\ kJ\ mol^{-1}$
  2. $0.632\ kJ\ mol^{-1}$
  3. $60.32\ kJ\ mol^{-1}$
  4. $-10.632\ kJ\ mol^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ \triangle G }^{ 0 }=-RTlnK$             $T={ 27 }^{ 0 }C=300K$

${ \triangle G }^{ 0 }=-8.3\times 300\times ln73$
          $=-10.632KJ{ mol }^{ -1 }$

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

The Van't Hoff equation is :

  1. $\Delta G^{\circ} = RT log _e K _p$
  2. $-\Delta G^{\circ} = RT log _e K _p$
  3. $\Delta G^{\circ} = RT^2 lnK _p$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Van't Hoff equation gives the relationship between the standard gibbs free energy change and the equilibrium constant.  It is represented by the equation $-\Delta  G^{\circ} = RT    log _e   K _p$.

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

if for the heterogeneous equilibrium $CaCO _{3}(s)\rightleftharpoons CaO(s)+CO _{2}(g);$ K=1 at 1 atm, the temperature is given by:

  1. $T=\frac{\Delta S^{0}}{\Delta H^{0}}$
  2. $T=\frac{\Delta H^{0}}{\Delta S^{0}}$
  3. $T=\frac{\Delta G^{0}}{ R^{0}}$
  4. $T=\frac{\Delta G^{0}}{\Delta H^{0}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta G = 2.303RT\space logK$


As K =1 , $\Delta G = 0$

We know the relation,

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

$T = \dfrac{\Delta H}{\Delta S}$

Option B is correct

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

Vant Hoff's equation is ___.

  1. ${log\frac{K _2}{K _1}=\frac{-\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2T _1} \right ]}$.
  2. ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2+T _1} \right ]}$.
  3. ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2T _1} \right ]}$.
  4. ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2+T _1}{T _2T _1} \right ]}$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The van't Hoff equation provides information about the temperature dependence of the equilibrium constant. The van't Hoff equation may be derived from the Gibbs-Helmholtz equation, which gives the temperature dependence of the Gibbs free energy.

The van't Hoff equation is $K=Ae^{\Delta H/RT}$ or $\displaystyle\frac {d ln K}{\partial T}=\frac {\Delta H}{RT^2}$

By, integrating the above equation, you will get the required relation.

Hence, the given statement is correct
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Calculate the standard voltage that can be obtained from an ethane oxygen fuel cell at $25^o C$.
$C _2H _6(g) + 7/2O _2(g) \rightarrow 2CO _2(g) + 3H _2O(1); \Delta G^o = -1467 \,kJ$

  1. $+0.91$
  2. $+0.54$
  3. $+0.72$
  4. $+1.08$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The standard cell potential is calculated using E_cell = -Delta G^o / (n F). For the ethane-oxygen fuel cell, burning one mole of C2H6 involves 14 electrons (since carbon goes from -3 to +4, 2 carbons give 14 electrons). Delta G^o = -1467 kJ = -1467000 J. F = 96500 C/mol. E = 1467000 / (14 * 96500) = +1.08 V.

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

Expansion of a perfect gas into vacuum is related with:

  1. $\Delta H=0$
  2. $q=0$
  3. $W=0$

  4. All the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Solution -
If an ideal gas or perfect gas 
expands into vacuum, it does 
no work 
i.e, work done = 0 

& this process is considered to 
be an adiabatic process, 
where $ q = 0 $
$ \Delta U = 0 $

Also, $\Delta H = \Delta U+Work \,done $
$ \Delta H = 0+0 $
$ \Delta H = 0 $

Hence, the answer is all of these.
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