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

Calculate the Standard Free Energy Change at 25 degrees celsius given the Equilibrium constant of 1.3 x 10^4.

  1. +23.4 kJ
    • 3.22 x 10^4 kJ

  3. -23,400 kJ

  4. -23.4 kJ

  5. +23,400 kJ
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 The temperature is 25 deg C or 298 K.
$\displaystyle  \Delta G^o = -RT ln K = - 8.314 \times 298 \times ln 1.3 \times 10^4 = -23469 J = -23.4 kJ$
Hence, the standard free energy change is -23.4 kJ

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

The cell in which the following reaction occurs:
$2Fe^{3+} _{(aq)}+2I^- _{(aq)}\rightarrow 2Fe^{2+} _{(aq)}+I _{2(s)}$ has $E^o _{cell}=0.236\ V$ at $298\ K$.
The equilibrium constant of the cell reaction is:

  1. $6.69\times 10^{-7}$
  2. $7.69\times 10^{-7}$
  3. $9.69\times 10^7$
  4. $6.69\times 10^7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know
$\log K _c=\dfrac{nFE^0 _{cell}}{2.303RT}$
where,
$n=2$
$F=96487$
$E^0 _{cell}=0.236\ V$
$R=8.31$
$T=298$
Substituting the values, we get
$\log K _c=\dfrac{2\times96487\times0.236}{2.303\times8.31\times298}$
$\log K _c=7.9854$
$K _c=antilog (7.9854)$
$K _c=9.69\times10^7$
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 $\Delta G=0$.
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. -ve , >1 , -ve

  2. -ve , <1 , -ve

  3. +ve , >1 , -ve

  4. -ve , >1 , +ve
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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

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

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

  1. Zero
  2. $1$
  3. $2$
  4. $10$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

$\Delta G^o (298 K)$ 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
$N _2+3H _2\overset {K _2}{\rightleftharpoons} 2NH _3$
$NH _3\overset {K _3}{\rightleftharpoons } \dfrac12N _2+\dfrac32H _2$

  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 $
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $

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.