Tag: relationship between equilibrium constant, reaction quotient and gibbs energy

Questions Related to relationship between equilibrium constant, reaction quotient and gibbs energy

Hydrolysis of sucrose gives glucose and fructose. The reaction takes place as: Sucrose $+ H _{2}O \rightleftharpoons$ Glucose $+$ Fructose. The equilibrium constant $K _{c}$ for this reaction is $2\times 10^{13}$ at $300\ K$. The $ \Delta G^{\circ}$ at $300\ K$ is:

chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

  1. $7.64\times 10^{4}\ J\ mol^{-1}$

  2. $7.64\times 10^{-4}\ J\ mol^{-1}$

  3. $-7.64\times 10^{-4}\ J\ mol^{-1}$

  4. $-7.64\times 10^{4}\ J\ mol^{-1}$

Show answer
Correct Option: D
Explanation:

The relationship between the standard free energy change $(\Delta G^o)$ and the equilibrium constant $(K _p)$ is $(\Delta G^o)=-RTlnK _c$, where, $R$ is the ideal gas constant and $T$ is the temperature.


Given, $K _c=2\times 10^{13}, T=300K, R=8.314 :Jmol^{-1}K^{-1}$

Substituting these values in the above expression, we get

$(\Delta G^o)=-RTlnK _c=-8.314 \times 300 \times ln(2\times 10^{13})=-7.64\times 10^{4} J mol^{-1}$ 

Hence, the standard free energy change $(\Delta G^o)=-7.64\times 10^{4} J mol^{-1}$

 $SO _2(g) + 1/2O _2 (g)\rightleftharpoons SO _3(g) \Delta H^o _{298} = 98.32 kJ/mole, \Delta S^o _{298} = 95.0 J/K/mole$.


 Find the $K _p$ for this above reaction at 298K:

chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

  1. $K _P = 9.31 \times 10^{-12} atm^{1/2}$

  2. $K _P = 5.34 \times 10^{-13} atm^{1/2}$

  3. $K _P = 3.7 \times 10^{-13} atm^{1/2}$

  4. $K _P = 3.7 \times 10^{-14} atm^{1/2}$

Show answer
Correct Option: B
Explanation:
$\displaystyle \Delta G^0 = \Delta H^0 - T\Delta S^0  $
$\displaystyle  \Delta G^0 =  98.32 \times 1000 - 298 \times 95.0 = 70010 J/mol$
$\displaystyle  \Delta G^0 = -RTlnK _P$
$\displaystyle 70010 = - 8.314 \times 298 \times ln K $
$\displaystyle  ln K = -28.26$
$\displaystyle  K = 5.34 \times 10^{-13}$