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

For the reaction at $298 K$


$A (g) + B (g)\rightleftharpoons C (g) + D (g)$

$\Delta H^o = 29.8 kcal ; \Delta S^o = 0.1 kcal/K$

Calculate $\Delta G^o$ and $K$.

  1. $\Delta G^o = 0 ; K = 1$
  2. $\Delta G^o = 1 ; K = e$
  3. $\Delta G^o = 2 ; K = e^2$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As we know,


$\Delta G^o = \Delta H^o - T\Delta S^o$ 

         $= 29.8 - ( 298\times0.1 )$

         $= 29.8-29.8=0$

Therefore, $\Delta G^o = 0$

The relation between $\Delta G^0 $ and $K$

$\Delta G^0$ = $ - RT lnK$

$K = 1 $

So, the correct option is $A$

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

When $\displaystyle \Delta G$ is zero :

  1. reaction moves in forward direction

  2. reaction moves in backward direction

  3. system is at equilibrium

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When $\displaystyle \Delta G$ is zero, system is at equilibrium.
Positive free energy change corresponds to non spontaneous reaction.
Negative free energy change corresponds to spontaneous reaction.

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

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?

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

  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$^{ -1 }$.
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$\Delta _rH = 75.68 kJ mol^{1}$

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

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

  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. $K _P = e^{ -\Delta G^{ \ominus }/RT }$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

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

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

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

  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 $
Reveal answer Fill a bubble to check yourself
D Correct answer
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. $K={ 10 }^{ { -\Delta G }^{ 0 }/2.303RT }$

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

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:

  1. $\Delta { H }^{ o }/RT$
  2. $-\Delta { H }^{ o }/R$
  3. $\Delta { H }^{ o }/R$
  4. $R/\Delta { H }^{ o }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Van't Hoff equation is ln(K) = -deltaH/R * (1/T) + constant. Plotting ln(K) vs 1/T results in a straight line with a slope of -deltaH/R.

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

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, $\displaystyle { K } _{ eq }$

  1. I and II only

  2. I and III only

  3. III only

  4. I, II, III

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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

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:

  1. 12 kJ/mole

  2. 831.4 kJ/mole

  3. 100 kJ/mole

  4. 88.57 kJ/mole
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The fraction of reactant molecules in the activated state represents the Boltzmann factor or can be related via the Arrhenius rate expression. Given that 3.8 * 10^-14 percent (or 3.8 * 10^-16 fraction) exist in the activated state, we use the relation exp(-Ea / (RT)) = 3.8 * 10^-16. Taking the natural logarithm: -Ea / (RT) = ln(3.8 * 10^-16) = -36.2. With T = 300 K and R = 8.314 J/mol K, solving for Ea yields approximately 100 kJ/mol.

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

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

  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) } $
Reveal answer Fill a bubble to check yourself
C Correct answer
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.