For the Haber’s process for the formation of ${\rm{N}}{{\rm{H}}_3}$ at ${\rm{298}}\;{\rm{K}}$ is : ${{\rm{N}}_2} + 3{{\rm{H}}_2}\;\leftrightharpoons2{\rm{N}}{{\rm{H}}_3};\;\;\Delta H = - 460\;\;{\rm{kJ}}$ Which of the following is correct?

The condition for equilibrium is ${{\rm{G}}_{{{\rm{N}}_2}}} + 3{{\rm{G}}_{{{\rm{H}}_2}}} = 2{{\rm{G}}_{{\rm{N}}{{\rm{H}}_3}}},$ where ${\rm{G}}$ is Gibbs energy per mole of gaseous species measured at that partial pressure

On addition ${{\rm{N}}_2}$, the equilibrium will shift to forward direction because according to II law of thermodynamics the entropy must decrease in the direction of spontaneous reaction.

The catalyst will increase the rate of forward reaction by 2 times and that of backward reaction by 1.5 times

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JEE CHEMISTRY MCQ: Equilibrium

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For the Haber’s process for the formation of

\mathrm{N}{\mathrm{H}_3}

\mathrm{298}\;\mathrm{K}

is :

{\mathrm{N}_2} + 3{\mathrm{H}_2}\;\leftrightharpoons2\mathrm{N}{\mathrm{H}_3};\;\;\Delta H = - 460\;\;\mathrm{kJ}

Which of the following is correct?

View answer and explanation

Correct option: A

The condition for equilibrium is

{\mathrm{G}_{{\mathrm{N}_2}}} + 3{\mathrm{G}_{{\mathrm{H}_2}}} = 2{\mathrm{G}_{\mathrm{N}{\mathrm{H}_3}}},

where

\mathrm{G}

is Gibbs energy per mole of gaseous species measured at that partial pressure

Option A (The condition for equilibrium is

{\mathrm{G}_{{\mathrm{N}_2}}} + 3{\mathrm{G}_{{\mathrm{H}_2}}} = 2{\mathrm{G}_{\mathrm{N}{\mathrm{H}_3}}}}

, where

\mathrm{G}

is Gibbs energy per mole of gaseous species measured at that partial pressure) is correct because at equilibrium, the change in Gibbs free energy is zero, meaning the Gibbs free energy of the reactants equals the Gibbs free energy of the products.

JEE CHEMISTRY MCQ: Equilibrium

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