Tag: law of mass action

Questions Related to law of mass action

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

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

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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 $
$\displaystyle  ln K = -28.26$
$\displaystyle  K = 5.34 \times 10^{-13}$