Tag: gibbs free energy

Questions Related to gibbs free energy

Calculate the equilibrium constant at 25 degrees celsius given the Standard Free Energy value of - 107.2 kJ

chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity
    • 43.2

  2. 43.2

  3. 6.18 x $ 10^8$

  4. 1.04

  5. 6.18 x $10^9$

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Correct Option: C
Explanation:

 The temperature is 25 deg C or 298 K
$\displaystyle  \Delta G^0 = -RT lnK$
$\displaystyle  \Delta G^0 = - 107.2 kJ = -107200 J$
$\displaystyle  -107200 = - 8.314 \times 298 \times ln K$
$\displaystyle  ln K = 43.268$
$\displaystyle  K = 6.18 \times 10^{18}$

A large positive value of $\Delta { G }^{ o }$ corresponds to which of these?

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

  1. Small positive $K$

  2. Small negative $K$

  3. Large positive $K$

  4. Large negative $K$

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Correct Option: A
Explanation:

A large positive value of $\Delta { G }^{ o }$ corresponds to small positive $K$.
$\Delta { G }^{ o }=-2.303RT\log { { K } _{ c } } $
When $ \displaystyle  { K } _{ c }  >0$, $ \displaystyle \Delta { G }^{ o } <0 $ and vice versa.

If $\Delta G$ standard is zero, this means :

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

  1. the reaction is both spontaneous and at equilibrium

  2. the system is at equilibrium at standard conditions

  3. the reaction is non spontaneous at standard conditions

  4. the reaction is spontaneous at standard conditions

  5. the reaction is both non spontaneous and at equilibrium

Show answer
Correct Option: B
Explanation:

If $\Delta G = 0$ this means, the system is at equilibrium at standard conditions.
$\Delta G = 0$, it means the reaction is equilibrium at standard conditions.
A negative value of $\Delta G  $, means spontaneous.
A positive value of $\Delta G $, means non-spontaneous.

If ${E} _{cell}^{o}$ for a given reaction is negative, which gives the correct relationships for the values of $\Delta { G }^{ o }$ and ${K} _{eq.}$?

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

  1. $\Delta { G }^{ o }>0,{ K } _{ eq. }<1$

  2. $\Delta { G }^{ o }>0,{ K } _{ eq. }>1$

  3. $\Delta { G }^{ o }<0,{ K } _{ eq. }>1$

  4. $\Delta { G }^{ o }<0,{ K } _{ eq. }<1$

Show answer
Correct Option: A
Explanation:
$\mathbf{Explanation:}$

From the relation between change in free energy$(\Delta G)$ and equilibrium constant $(K _{eq})$ we have :

$\mathbf{\Delta G=-RTlnK _{eq}}$      $\mathbf{\rightarrow (1)}$

where:
$\Delta G=$ The change in free energy
$R=$ Gas constant
$T=$ The absolute temperature
$K _{eq}=$ Equilibrium constant
 
From the relation between change in energy $\Delta G$ and $E _{cell}(E^{o})$ we have :

$\mathbf{\Delta G=-nFE^{o}}$ $\mathbf{\rightarrow (2)}$

$\Delta G=$ The change in free energy
$E^{o}=E _{cell}$
$n =$ moles of e- from balanced redox reaction
$F =$ Faraday's constant 

From equation  $(2) $ if $E^{o}< 0$ then $\Delta G>0$$\mathbf{\rightarrow (3)}$

If then $lnK<1$ and then $K _{eq}<1$ that is positive $\mathbf{\rightarrow (4)}$ 

From $(3)$ and $(4)$ we get that

$\mathbf{\Delta G>0}$ and $\mathbf{K _{eq}}<1$

Hence the correct answer is option $A$.

Consider the reaction of extraction of gold from its ore
$Au + 2CN^{-} (aq.) + \dfrac {1}{4}O _{2}(g) + \dfrac {1}{2}H _{2}O\rightarrow Au(CN) _{2}^{-} + OH^{-}$
Use the following data to calculate $\triangle G^{\circ}$ for the reaction
$K _{f} \left {Au(CN) _{2}^{-}\right ) = X$
$O _{2} + 2H _{2}O + 4e^{-}\rightarrow 4OH^{-}; E^{\circ} = +0.41\ volt$
$Au^{3+} + 3e^{-}\rightarrow Au; E^{\circ} = + 1.5\ volt$
$Au^{3+} + 2e^{-} \rightarrow Au^{+}; E^{\circ} = + 1.4\ volt$.

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

  1. $-RT\ ln\ X + 1.29\ F$

  2. $-RT\ ln\ X - 2.11\ F$

  3. $-RT\ ln \dfrac {1}{X} + 2.11\ f$

  4. $-RT\ ln\ X - 1.29\ F$

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Correct Option: A

The value of $log _{10}$ K for a reaction $A\rightleftharpoons B$ is:

$( Given : \Delta _{r}H^{0} _{298k}=-54.07 kJ mol^{-1},$ $\Delta _{r}S^{0} _{298k}=10JK^{-1}mol^{-1}$ $and\ R=8.314 JK^{-1}mol^{-1};$ 
$2.303\times 8.314\times 298=5705 )$

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

  1. 5

  2. 10

  3. 95

  4. 100

Show answer
Correct Option: B
Explanation:

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}=-54.07 \times 1000 - 298 \times 10$

$=-54070-2980=-57050$

$\Delta G^{0}=-2.303 RT log _{10}K$

$-57050=-2.303\times 298\times 8.314 log _{10}K=-5705 log _{10}K$


$ log _{10}K=10$