Tag: energetics and thermochemistry

Questions Related to energetics and thermochemistry

The equilibrium constant of a reaction is 10. What will be the value of $\Delta G^0$ at 300 K?

chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity
    • 5.74 kJ
    • 574 kJ
    • 11.48 kJ

  4. +5.74 kJ

Show answer
Correct Option: A
Explanation:

$\Delta G $= -2.303 RT log K
 = $-2.303 \times 8.314 \times 300 \ log 10$
=-5.74 kJ

A reaction attains equilibrium state under standard conditions. Identify the incorrect option regarding this statement.

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

  1. Equilibrium constant K = 0

  2. Equilibrium constant K = 1

  3. $\Delta G^0$ = 0 and $\Delta H^0$ = T$\Delta S^0$

  4. All options are correct

Show answer
Correct Option: A
Explanation:

$\Delta G^0 = 0 $ at equilibrium under standard state.
Also at equilibrium, $\Delta G  = 0 $
$\therefore$ $\Delta H^0 - T\Delta S^0 = 0$
Also $\Delta G^0 $= -2.303 RT Iog K  
$\therefore$ K = 1

For a spontaneous reaction the $\Delta G$, equilibrium constant $(K _{eq})$ and $E^{0} _{cell}$ will be respectively 

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

  1. -ve , >1 , -ve

  2. -ve , <1 , -ve

  3. +ve , >1 , -ve

  4. -ve , >1 , +ve

Show answer
Correct Option: D
Explanation:

$\Delta G^{o}=- R TlnKeq$

$\Delta G^{o}=$ will be negative for spontaneous process

$Keqm > 1$for spontaneous process 

$E^{o} cell > O$ for spontaneous process

$nFE _{cell}^{o}= RT ln Keq$

$\therefore E _{cell}^{o}>o$

For a reversible reaction, if $\Delta { G }^{ o }=0$, the equilibrium constant of the reaction should be equal to:

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

  1. Zero

  2. $1$

  3. $2$

  4. $10$

Show answer
Correct Option: B
Explanation:
If $\Delta {G}^{o}=0$

At equilibrium $\Delta G=0$

$\Delta G=\Delta {G}^{o}+RT\ln {K} _{eq}$

$0=0+RT\ln {K} _{eq}$

$\ln {K} _{eq}=0$

${K} _{eq}=1$

equilibrium constant $=1$

Option B is correct.

$\Delta G^o (298 K)$ for the reaction $\dfrac12 N _2+\dfrac32H _2\overset {K _1}{\rightleftharpoons} NH _3$ is -16.5 kJ $mol^{-1}$. The equilibrium constant $(K _1)$ at $25^oC$ & the equilibrium constant $K _2$ and $K _3$ for the following reactions are
$N _2+3H _2\overset {K _2}{\rightleftharpoons} 2NH _3$
$NH _3\overset {K _3}{\rightleftharpoons } \dfrac12N _2+\dfrac32H _2$

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

  1. $K _1 = 779.4, K _2 = 6.074 \times 10^{5} ; K _3 = 1.283 \times 10^{-3}$

  2. $K _1 = 779.4, K _2 = 2.183 \times 10^{5} ; K _3 = 3.576 \times 10^{3}$

  3. $K _1 = 124.4, K _2 = 6.074 \times 10^{5} ; K _3 = 2.34\times 10^{3}$

  4. $None ::of ::these $

Show answer
Correct Option: A
Explanation:


 $\displaystyle \Delta G^o = -RTlnK _1$
 $\displaystyle -16500 = - 8.314 \times 298 \times lnK _1$
$\displaystyle 6.6597 = ln K _1$
 $\displaystyle K _1 = 779.4$
$\displaystyle K _2 = K _1^2 = (779.4)^2 = 6.074 \times 10^5$
 $\displaystyle K _3 = \dfrac {1}{K _1}=\dfrac {1}{779.4}=1.283 \times 10^{-3} $

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 $10^9$

Show answer
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 $K$

Show answer
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. $-RT\ ln\ X - 1.29\ F$

Show answer
Correct Option: A