Tag: energetics and thermochemistry

Questions Related to energetics and thermochemistry

The density of an equilibrium mixture of $N _2O _4$ and $NO _2$ at 101.32 $KP _a$ is 3.62 g $dm^{3}$ at 288 K and 1.84 g $dm^{3}$ at 348 K. 


What is the heat of the reaction for the following reaction?

$N _2O _4\rightleftharpoons 2NO _2(g)$

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

  1. $\Delta _rH = 37.29 $  kJ mol$^{ -1 }$. 

  2. $\Delta _rH = 75.68 $  kJ mol$^{ -1 }$. 

  3. $\Delta _rH = 95.7$  kJ mol$^{ -1 }$. 

  4. $\Delta _rH = 151.3 $  kJ mol$^{ -1 }$. 

Show answer
Correct Option: B
Explanation:

At 288 K, $M _{avg.}=\frac {3.62\times 0.0821\times 288}{1}$
$\frac {92}{M _{avg.}}=1+\alpha \Rightarrow K _{P1}=\frac {4\alpha^2}{1-\alpha^2}$
Similarly at $348 K, M'/avg.=\frac {1.84\times 0.0821\times 348}{1}$
$\frac {92}{M'avg}=1+\alpha'\Rightarrow K _{P _2}=\frac {4\alpha'^2}{1-\alpha'^2}$
$log \frac {K _{P _2}}{K _{P _1}}=\frac {\Delta H^o}{2.303 R}\left [\frac {1}{288}-\frac {1}{348}\right ]$
so,
$\Delta _rH = 75.68 kJ mol^{1}$

Which is not correct relationship between $\Delta G^{ \ominus }$ and equilibrium constant $K _P$

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

  1. $K _P = -RT log \Delta G^{ \ominus }$

  2. $K _P = [e/RT]^{ \Delta G^{ \ominus } }$

  3. $K _P = -\frac { \Delta G^{ \ominus } }{ RT }$

  4. $K _P = e^{ -\Delta G^{ \ominus }/RT }$

Show answer
Correct Option: A,B,C
Explanation:

$\Delta G = \Delta G^{ \ominus } + RT log K$
and at equilibrium,
$\Delta G = 0$ so
$\Delta G^{ \ominus } = - RT log K$
$\Delta G^{ \ominus } = -RTln K _P$
$K _P = e^{ -\Delta G^{ \ominus }/RT }$

The correct relation between equilibrium constant $(K)$, standard free  energy $(\Delta {G}^{o})$ and temperature $(T)$ is:

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

  1. $\Delta {G}^{o}=RT\ln {K}$

  2. $K={ e }^{ \Delta { G }^{ o }/2.303 RT }\quad $

  3. $\Delta { G }^{ o }=-RT\log{K}$

  4. $K={ 10 }^{ -\Delta { G }^{ o }/2.303 RT }\quad $

Show answer
Correct Option: D
Explanation:

Consider a reaction, $A+B\rightleftharpoons C+D$

$\Delta G={ \Delta G }^{ 0 }+RTlnQ$
for equilibrium, $\Delta G=0$
$\therefore \quad 0={ \Delta G }^{ 0 }+RTlnK$
$\therefore \quad { \Delta G }^{ 0 }=-RTlnK$
$\therefore \quad { \Delta G }^{ 0 }=-2.303RTlogK$
i.e. $K={ 10 }^{ { -\Delta G }^{ 0 }/2.303RT }$

When $\ln{K}$ is plotted against $\cfrac { 1 }{ T } $ using the Van't Hoff equation, a straight line is expected with a slope equal to:

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

  1. $\Delta { H }^{ o }/RT$

  2. $-\Delta { H }^{ o }/R$

  3. $\Delta { H }^{ o }/R$

  4. $R/\Delta { H }^{ o }$

Show answer
Correct Option: B

If we know $\displaystyle { \Delta G }^{ \circ  }$ of a reaction, which of the following can be defined ?
I. Cell potential, $\displaystyle { E }^{ \circ  }$
II. Activation energy, $\displaystyle { E } _{ a }$
III. Equilibrium constant, $\displaystyle { K } _{ eq }$

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

  1. I and II only

  2. I and III only

  3. III only

  4. I, II, III

  5. None of these

Show answer
Correct Option: B
Explanation:

If we know $\displaystyle \Delta G^o $  of a reaction, the following can be defined. 
I. Cell potential, $\displaystyle E^o $
III. Equilibrium constant, Keq
$\displaystyle \Delta G^o = -nF E^o = - RTlnK $

For the first order reaction $A\longrightarrow B+C$, carried out at $27^0C  $ if  $ 3.8\ \times \ 10^{ -16 } \%$ of the reactant molecules exists in the activated state, the ${ E } _{ a }$ (activation energy) of the reaction is:

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

  1. 12 kJ/mole

  2. 831.4 kJ/mole

  3. 100 kJ/mole

  4. 88.57 kJ/mole

Show answer
Correct Option: A

By which of the following relations, the equilibrium constant varies with temperature?

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

  1. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =\cfrac { \Delta { H }^{ o } }{ R } \int _{ { T } _{ 1 } }^{ { T } _{ 2 } }{ d\left( \cfrac { 1 }{ T } \right) } $

  2. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =-\cfrac { \Delta { H }^{ o } }{ R } \int _{ { 1/T } _{ 1 } }^{ { 1/T } _{ 2 } }{ d\left( \cfrac { 1 }{ { T }^{ 2 } } \right) } $

  3. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =-\cfrac { \Delta { H }^{ o } }{ R } \int _{ { T } _{ 1 } }^{ { T } _{ 2 } }{ d\left( \cfrac { 1 }{ T } \right) } $

  4. $\ln { { K } _{ 2 } } -\ln { { K } _{ 1 } } =-\cfrac { \Delta { H }^{ o } }{ R } \int _{ { 1/T } _{ 2 } }^{ { 1/T } _{ 1 } }{ d\left( \cfrac { 1 }{ { T }^{ } } \right) } $

Show answer
Correct Option: C
Explanation:
$\textbf{Explanation:}$

  • We can use $\mathit{Gibbs-Helmholtz}$ to get the temperature dependence of $K$                                                                                            
$\mathbf{\left ( \frac{\partial \left [ \Delta _{r}G^{o} \right ]}{\partial T} \right )}$   $\mathbf{=\frac{-\Delta _{r}H^{o}}{T^{2}}}$     $\mathbf{\rightarrow \left ( 1 \right )}$

  • At equilibrium, we can equate $\Delta _{r}G^{o}$ to $-RTlnK$ so we get

  $\mathbf{\left ( \frac{\partial \left [ lnK \right ]}{\partial T} \right )= \frac{\Delta _{r}H^{o}}{RT^{2}}}$    $\mathbf{\rightarrow (2)}$

  • We see that whether  K  increases or decreases with temperature is linked to whether the reaction enthalpy is positive or negative. If the temperature is changed little enough that  $\Delta _rH^{o}$  can be considered constant, we can translate a  $K$  value at one temperature into another by integrating the above expression, we get a similar derivation as with melting point depression:

$\mathbf{ln\frac{K\left ( T _{2} \right )}{K\left ( T _{1} \right )}=\frac{-\Delta _{r}H^{o}}{R}\left ( \frac{1}{T _{2}}-\frac{1}{T _{1}} \right )}$$\mathbf{\rightarrow \left ( 3 \right )}$

  • If we integrate and differentiate the left side of the equation then we get and solve the right side we get

$\mathbf{lnK _{2}-lnK _{1}=\frac{-\Delta H^{0}}{R}\int _{T _{1}}^{T _{2}}\mathbf{\mathit{d}}\left ( \frac{1}{T} \right )}$$\mathbf{\rightarrow (4)}$

Hence from equation $4$ we can say that option $C$ is correct.

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 $\Delta G=0$.