Tag: chemical equilibrium and acids-bases

Questions Related to chemical equilibrium and acids-bases

In an equilibrium reaction for which $\Delta G^{\circ}$ = 0. Determine the value of equilibrium constant K. $\delta G$ =0

  1. 1.0

  2. 2.7

  3. 7.6

  4. 8.5


Correct Option: A
Explanation:

$\triangle G=-RTlnK\ If\quad \triangle G=0,E=0,Q=K\ No\quad reaction\quad takes\quad place$

$K {c}$ for an equilibrium $SO _{3}\rightleftharpoons SO _{2}(g)+ \frac{1}{2}O _{2}(g)$ is equal to 0.15 at 900 K. The equilibrium constant for the equation $2SO _{2}+ O _{2}\rightleftharpoons 2SO _{3}(g)$ is ___________.

  1. $6.66$

  2. $44.4$

  3. $2.25$

  4. $0.44$


Correct Option: B
Explanation:

$K _{c}$ for $SO _{3}\rightleftharpoons  SO _{2}+ \frac{1}{2} O _{2}$ is 0.15.

$K _{c}$ for the following reaction $ 2SO _{2}+ O _{2}\rightleftharpoons 2SO _{3}$ will be the reciprocal of square of the above equilibrium as stoichiometric coefficients are doubled and the reaction is reversed.

$K^{1} _{c} = \dfrac{1}{K^{2} _{c}} = \dfrac{1}{0.15^{2}} = 44.4$

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $K$ is:

  1. $-\Delta G^{\circ}\, =\, RT\, in \, K$

  2. $\Delta G\, =\, RT\, in\, K$

  3. $-\, \Delta G \, =\, RT\, in\, K$

  4. $\Delta G^{\circ}\, =\, RT\, in\, K$


Correct Option: A
Explanation:

The equilibrium constant is related to the free energy change of the reaction by the expression:

$K = e^{(-\Delta G^o/RT)}$ or $ln K = - \Delta G^o/RT$

or  $- \Delta G^o= RT \ ln K$ in which $T$ is the temperature in Kelvin and $R$ is the gas constant (1.986 cal/K mol)

Calculate value of $'ln(K _{eq})$' for the reaction at 250 K.
$N _2O _4 (g) \rightleftharpoons 2NO _2 (g)$
Given: $H^0 _f((NO _2)g) = + 40.407 kJ / mol$
$H^0 _f((N _2O _4)g) = + 70 kJ / mol$
$S^0 _r = 10 JK^{-1}$

  1. 4

  2. -4

  3. 1.2

  4. -1.2


Correct Option: B
Explanation:

$+G^0 = +H^0 -TS^0$
$+H _r^0 = (+ H _f^0)p - ( + H _f^0) _R$
$= 2 \times (+ 40.407) - (+70)$
$= + 10.814 kJ$
$+G^0 = + 10.814 kJ - 250 \times 10 J/K$
$= + 10.814 kJ - 2500 J = 8314 J$
$+G^0 = -RT ln K$
$8314 = - 8.314 \times 250 ln K$
$ln K = - 4.$

Assertion: For every chemical reaction at equilibrium, the standard Gibbs energy of reaction is zero.
Reason: At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

  1. The Assertion is True, Reason is True; Reason is the correct explanation for Assertion

  2. The Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion

  3. The Assertion is True, Reason is False

  4. The Assertion is False, Reason is True


Correct Option: D
Explanation:


The standard Gibb's energy for a reaction is given by $\Delta G^o$.

At equilibrium, $\Delta \mathrm{G}=0$ whereas $\ \Delta \mathrm{G}^{\mathrm{o}}$ for a reaction may or may not be zero.
For a spontaneous process, Gibb's energy for a reaction is always negative, $\Delta \mathrm{G}<0$
Hence, Assertion is false but Reason is true.

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant, $K$ is:

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

  2. $\Delta G = RT ln K$

  3. $\Delta G = -RT ln K$

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


Correct Option: A
Explanation:

The correct relationship between free energy change $\Delta G^{o}$ in a reaction and the corresponding equilibrium constant $K$ is $\Delta G^{o} = -RT :ln K$.

Here, $R$ is the ideal gas constant and $T$ is the temperature.

The value of equilibrium constant $(K _f)$ for the reaction: $Zn^{2+}(aq)+4OH^{-}(aq)\rightleftharpoons Zn(OH) _{4}^{2-}(aq)$ is represented in scientific notation as $p\times10^{q},$ then q is:
Given : $Zn^{2+}(aq)+2e^{-}\rightarrow Zn(s);  E^{0}=-0.76 V$
            $Zn(OH) _{4}^{2-}(aq)+2e^{-}\rightarrow Zn(s)+4OH^{-}(aq);  E^{0}=-1.36V$
            $2.303\dfrac{RT}{F}=0.06$

  1. 20

  2. 10

  3. 15

  4. 21


Correct Option: A
Explanation:

$\Delta G^o=-RTlnK _{eq} $

$logK _{eq}=\dfrac{nFE^{o}}{RT\times 2.303}\Rightarrow \dfrac{2\times 0.6}{0.06}\Rightarrow 20$

$K=10^{20}$


Greenhouse gas $CO _2$ can be converted to $CO(g)$ by the following reaction

$CO _2(g)+H _2(g)\rightarrow CO(g)+H _2O(g)$,
 
termed as water gas reaction.

The Equilibrium constant $K _p$ for the water gas reaction at $1000\;K$ is: 

 $(\Delta H _{\displaystyle1000\;K}=35040\;J\;mol^{-1}\ ; \Delta S _{1000\;K}=32.11\;J\;mol^{-1}\ K^{-1})$

(Note : The gases behave ideally).

  1. $K _p=0.7030$

  2. $K _p=0.7300$

  3. $K _p=0.7330$

  4. $K _p=0.7303$


Correct Option: A
Greenhouse gas $CO _2$ can be converted to $CO(g)$ by the following reaction

$CO _2(g)+H _2(g)\rightarrow CO(g)+H _2O(g)$,
 
termed as water gas reaction.

$\Delta G$ for the reaction at 1000 K is : 

 $(\Delta H _{\displaystyle1000\;K}=35040\;J\;mol^{-1}\ ; \Delta S _{1000\;K}=32.11\;J\;mol^{-1}\ K^{-1})$

  1. $2930\;J$

  2. $2934\;J$

  3. $2300\;J$

  4. None of these


Correct Option: A
Explanation:

A.2930J
$\triangle G=\triangle H-T\triangle S\ \triangle G=35040-1000\times 32.11\ \quad \quad \quad \quad =35040-32110\ \quad \quad \quad \quad =2930J$

Steam undergoes decomposition at high temperature as per the reaction
$H _{2}O(g) \rightleftharpoons  H _{2}(g)+\frac{1}{2}O _{2}(g), \Delta H^{\circ}=200 kJ  mol^{-1} \Delta S^{\circ}=40  J  mol^{-1}$
The temperature at which equilibrium constant is unit is :

  1. 3000 Kelvin

  2. 5000 Kelvin

  3. 5333 Kelvin

  4. 5 Kelvin


Correct Option: B
Explanation:

$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}=-RT  ln Keq = 0  [Keq = 1]$.
$\Rightarrow T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{200\times 10^{3}}{40}=5000  K$