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For the equilibrium, 

2H_{2}O \rightleftharpoons H_{3}O^{+}+OH^{-}, the value of \Delta G^{0} at 298 K is approximately :

  • Option 1)

    80 kJ mol -1

  • Option 2)

    -100 kJ mol -1

  • Option 3)

    -80 kJ mol -1

  • Option 4)

    100 kJ mol-1

Answers (1)

best_answer

 

Δ G of equilibrium -

\Delta G_{0}= -2.303 RT \log K_{c}
 

- wherein

At Equilibrum    

\Delta G= 0

and Q= K_{c}

 

 

 

ΔG{reaction} -

\Delta G _{reaction }= \sum \Delta G_{product}- \sum \Delta G_{reactant}
 

- wherein

\sum \Delta G_{product}=Sum of   \Delta G of all product

\sum \Delta G_{reactant}=Sum of   \Delta G of all reactant

 

\Delta G^{0}=-2303\: RT\: log K

=-2303\times 8.314\times 298\: log(10^{-14})

=+80\frac{KJ}{mol}

 


Option 1)

80 kJ mol -1

Option 2)

-100 kJ mol -1

Option 3)

-80 kJ mol -1

Option 4)

100 kJ mol-1

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