Tag: energetics and thermochemistry

Quizzes (6) Questions (104)

Questions Related to energetics and thermochemistry

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

Vant Hoff's equation is ___.

  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. ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2+T _1}{T _2T _1} \right ]}$.
Reveal answer Fill a bubble to check yourself
C Correct answer
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
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

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

  1. $+0.91$
  2. $+0.54$
  3. $+0.72$
  4. $+1.08$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The standard cell potential is calculated using E_cell = -Delta G^o / (n F). For the ethane-oxygen fuel cell, burning one mole of C2H6 involves 14 electrons (since carbon goes from -3 to +4, 2 carbons give 14 electrons). Delta G^o = -1467 kJ = -1467000 J. F = 96500 C/mol. E = 1467000 / (14 * 96500) = +1.08 V.

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

Expansion of a perfect gas into vacuum is related with:

  1. $\Delta H=0$
  2. $q=0$
  3. $W=0$

  4. All the above
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Which are correct representation at equilibrium?

  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. $\displaystyle \frac { P }{ { P }^{ o } } ={ e }^{ { -\Delta H }/{ RT } }$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

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

  1. $\Delta $ G is positive
  2. $\Delta $ H is positive
  3. $\Delta $ S is positive
  4. $\Delta $ A is positive
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Although the dissolution of ammonium chloride is endothermic (Delta H > 0), it dissolves spontaneously because the increase in entropy (Delta S > 0) is sufficiently large to make the overall Gibbs free energy change negative (Delta G = Delta H - T * Delta S < 0) at room temperature.

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

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

  1. $\displaystyle { \Delta G }^{ o }=RTIn{ K } _{ c }$
  2. $\displaystyle -{ \Delta G }^{ o }=RTIn{ K } _{ c }$
  3. $\displaystyle { \Delta G }=RTIn{ K } _{ c }$
  4. $\displaystyle -{ \Delta G }=RTIn{ K } _{ c }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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:

  1. $\displaystyle 1.97\times { 10 }^{ -3 }$
  2. $\displaystyle 1.97\times { 10 }^{ -2 }$
  3. $\displaystyle 1.97\times { 10 }^{ -4 }$
  4. $\displaystyle 1.97\times { 10 }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

van't Hoff equation is

  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. $K=Ae^{\Delta H/RT}$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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 $\frac {d ln K}{\partial T}=\frac {\Delta H}{RT^2}$

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

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

  1. $3.86 kcal$
  2. $6.93 kcal$
  3. $1.68 kcal$
  4. $1.68\times 10^{-2} kcal$
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
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