Tag: gibbs energy change and equilibrium

Quizzes (0) Questions (47)

Questions Related to gibbs energy change and equilibrium

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

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

    • 43.2

  2. 43.2
  3. 6.18 x $ 10^8$

  4. 1.04
  5. 6.18 x $10^9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

  1. Small positive $K$
  2. Small negative $K$
  3. Large positive $K$
  4. Large negative $K$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. <font><font class="">the reaction is both spontaneous and at equilibrium</font></font>

  2. <font><font>the system is at equilibrium at standard conditions</font></font>

  3. <font><font class="">the reaction is non spontaneous at standard conditions</font></font>

  4. <font><font>the reaction is spontaneous at standard conditions</font></font>

  5. <font><font>the reaction is both non spontaneous and at equilibrium</font></font>
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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. $\Delta { G }^{ o }<0,{ K } _{ eq. }<1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A$.
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

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
$K _{f} \left {Au(CN) _{2}^{-}\right ) = X$
$O _{2} + 2H _{2}O + 4e^{-}\rightarrow 4OH^{-}; E^{\circ} = +0.41\ volt$
$Au^{3+} + 3e^{-}\rightarrow Au; E^{\circ} = + 1.5\ volt$
$Au^{3+} + 2e^{-} \rightarrow Au^{+}; E^{\circ} = + 1.4\ volt$.

  1. $-RT\ ln\ X + 1.29\ F$
  2. $-RT\ ln\ X - 2.11\ F$
  3. $-RT\ ln \dfrac {1}{X} + 2.11\ f$
  4. $-RT\ ln\ X - 1.29\ F$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The standard Gibbs free energy change for the reaction is calculated using the standard cell potential. The reaction involves oxidation of Au and reduction of O2. The expression involves the formation constant X and the standard potentials provided.

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

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

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

  1. 5

  2. 10

  3. 95

  4. 100
Reveal answer Fill a bubble to check yourself
B Correct answer
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$