Tag: chemistry

Questions Related to chemistry

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

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

  1. +23.4 kJ

    • 3.22 x 10^4 kJ

  3. -23,400 kJ

  4. -23.4 kJ

  5. +23,400 kJ

Show answer
Correct Option: D
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

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:

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

  1. $6.69\times 10^{-7}$

  2. $7.69\times 10^{-7}$

  3. $9.69\times 10^7$

  4. $6.69\times 10^7$

Show answer
Correct Option: C
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
$\log K _c=\dfrac{2\times96487\times0.236}{2.303\times8.31\times298}$
$\log K _c=7.9854$
$K _c=antilog (7.9854)$
$K _c=9.69\times10^7$

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 :

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

  1. -1

  2. +1

  3. +2

  4. 0

Show answer
Correct Option: D
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 $\Delta G=0$.

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.