Tag: energetics and thermochemistry

Questions Related to energetics and thermochemistry

A reaction attains equilibrium, when the free energy change is

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

  1. $1$

  2. $2$

  3. $3$

  4. $0$

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Correct Option: D
Explanation:

As we know,
At equilibrium $\Delta G=0$

Vant Hoff's equation is ___.

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

  1.  ${log\frac{K _2}{K _1}=\frac{-\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2T _1} \right ]}$.

  2.  ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2+T _1} \right ]}$.

  3.  ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2T _1} \right ]}$.

  4.  ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2+T _1}{T _2T _1} \right ]}$.

Show answer
Correct Option: C
Explanation:
The van't Hoff equation provides information about the temperature dependence of the equilibrium constant. The van't Hoff equation may be derived from the Gibbs-Helmholtz equation, which gives the temperature dependence of the Gibbs free energy.

The van't Hoff equation is $K=Ae^{\Delta H/RT}$ or $\displaystyle\frac {d ln K}{\partial T}=\frac {\Delta H}{RT^2}$

By, integrating the above equation, you will get the required relation.

Hence, the given statement is correct

Calculate the standard voltage that can be obtained from an ethane oxygen fuel cell at $25^o C$.
$C _2H _6(g) + 7/2O _2(g) \rightarrow 2CO _2(g) + 3H _2O(1); \Delta G^o = -1467 \,kJ$

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

  1. $+0.91$

  2. $+0.54$

  3. $+0.72$

  4. $+1.08$

Show answer
Correct Option: C

Expansion of a perfect gas into vacuum is related with:

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

  1. $\Delta H=0$

  2. $q=0$

  3. $W=0$

  4. All the above

Show answer
Correct Option: D
Explanation:
Solution -
If an ideal gas or perfect gas 
expands into vacuum, it does 
no work 
i.e, work done = 0 

& this process is considered to 
be an adiabatic process, 
where $ q = 0 $
$ \Delta U = 0 $

Also, $\Delta H = \Delta U+Work \,done $
$ \Delta H = 0+0 $
$ \Delta H = 0 $

Hence, the answer is all of these.

Which are correct representation at equilibrium?

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

  1. $\displaystyle p=\frac { eRT }{ N } $

  2. $\displaystyle K={ e }^{ { { -\Delta G }^{ o } }/{ RT } }$

  3. $\displaystyle \frac { { K } _{ 1 } }{ { K } _{ 2 } } ={ e }^{ { { -E } _{ a } }/{ RT } }$

  4. $\displaystyle \frac { P }{ { P }^{ o } } ={ e }^{ { -\Delta H }/{ RT } }$

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Correct Option: A,B,C,D

Although dissolution of $NH _{4}Cl$ in water is endothermic yet it dissolves because:

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

  1. $\Delta $ G is positive

  2. $\Delta $ H is positive

  3. $\Delta $ S is positive

  4. $\Delta $ A is positive

Show answer
Correct Option: B

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $\displaystyle { K } _{ c }$ is:

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

  1. $\displaystyle { \Delta G }^{ o }=RTIn{ K } _{ c }$

  2. $\displaystyle -{ \Delta G }^{ o }=RTIn{ K } _{ c }$

  3. $\displaystyle { \Delta G }=RTIn{ K } _{ c }$

  4. $\displaystyle -{ \Delta G }=RTIn{ K } _{ c }$

Show answer
Correct Option: B
Explanation:

$\displaystyle \because \quad \Delta G={ \Delta G }^{ o }+RTInQ$, where Q is reaction quotient at equilibrium, $\displaystyle \Delta G=0$ and $\displaystyle Q={ K } _{ c }$
$\displaystyle \therefore \quad -\Delta G=RTIn{ K } _{ c }$

For the reaction : $\displaystyle 2NOCl(g)\longrightarrow 2NO(g)+{ Cl } _{ 2 }(g)$, The equilibrium constant at 400K, if $\displaystyle { \Delta H }^{ o }=77.18kJ{ mol }^{ -1 }$ and $\displaystyle { \Delta S }^{ o }=0.122kJ{ K }^{ -1 }{ mol }^{ -1 }$ is:

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

  1. $\displaystyle 1.97\times { 10 }^{ -3 }$

  2. $\displaystyle 1.97\times { 10 }^{ -2 }$

  3. $\displaystyle 1.97\times { 10 }^{ -4 }$

  4. $\displaystyle 1.97\times { 10 }^{ -1 }$

Show answer
Correct Option: C
Explanation:
Given the reaction: $2NOCl(g)\rightarrow 2NO(g)+Cl _2(g)$

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

$\Delta G^o=77.18-400\times 0.122kJmol^{-1}$

$\Delta G^o=28.38\ kJmol^{-1}$

$K=e^{(\dfrac{-\Delta G^o}{RT})} $

$=1.97\times 10^{-4}$

Hence, option C is correct.

van't Hoff equation is

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

  1. $(d/dT) ln K=-\Delta H/RT^2$

  2. $(d/dT) ln K=+\Delta H/RT^2$

  3. $(d/dT) ln K=-\Delta H/RT$

  4. $K=Ae^{\Delta H/RT}$

Show answer
Correct Option: B,D
Explanation:

The van't Hoff equation provides information about the temperature dependence of the equilibrium constant. The van't Hoff equation may be derived from the Gibbs-Helmholtz equation, which gives the temperature dependence of the Gibbs free energy.
The van't Hoff equation is $K=Ae^{\Delta H/RT}$
or $\frac {d ln K}{\partial T}=\frac {\Delta H}{RT^2}$

The rate of disappearance of A at two temperatures is given by $A\rightleftharpoons B$
i. $\frac {-d[A]}{dt}=2\times 10^{-2}[A]-4\times 10^{-3}[B]$ at 300 K
ii. $\frac {-d[A]}{dt}=4\times 10^{-2}[A]-16\times 10^{-4}[B]$ at 300 K
From the given values of heat of reaction which are incorrect

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

  1. $3.86 kcal$

  2. $6.93 kcal$

  3. $1.68 kcal$

  4. $1.68\times 10^{-2} kcal$

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

$K _1=\frac {K _f}{K _b}=\frac {2\times 10^{-2}}{4\times 10^{-3}}=5$ at 300 K
$K _2=\frac {K _f}{K _b}=\frac {4\times 10^{-2}}{16\times 10^{-4}}=25$ at 400 K
$\therefore 2.303 log\frac {25}{55}=\frac {\Delta H}{2}\times \left [\frac {400-300}{400\times 300}\right ]$
or $\Delta H=3.85 kcal$

Hence, option A is correct and others are incorrect