Question
For the reaction, $N_{2}O_{4} \rightleftharpoons 2{NO}_{2}$, graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is $- 5.40\text{ }kJ{\text{ }mol}^{- 1}$.
B. As ${AG}^{\ominus}$ in graph is positive, $N_{2}O_{4}$ will not dissociate into ${NO}_{2}$ at all.
C. Reverse reaction will go to completion.
D. When 1 mole of $N_{2}O_{4}$ changes into equilibrium mixture, value of $\Delta G^{\ominus} = - 0.84\text{ }kJ{\text{ }mol}^{- 1}$
E. When 2 moles of ${NO}_{2}$, changes into equilibrium mixture, $\Delta G^{\ominus}$ for equilibrium mixture is $- 6.24\text{ }kJ{\text{ }mol}^{- 1}$.
Choose the correct answer from the options given below:
Options
Solution
(A) $\Delta_{r}G^{\circ} = G^{\circ}\ _{B} - G^{\circ}\ _{A} = + ve$
(B) $\Delta_{r}G^{\circ} = + ve,N_{2}O_{4}$ will partially dissociates into ${NO}_{2}$.
(C) For reverse reaction
It is partially completed as there is equilibrium at E .
(D) For 1 mole $N_{2}O_{4};\Delta_{r}G^{\circ} = - 0.84\text{ }kJ{\text{ }mol}^{- 1}$
(E) For 2 mole ${NO}_{2};\Delta G^{\circ} = - 5.4 - 0.84$
$$= - 6.24\text{ }kJ{\text{ }mol}^{- 1}$$
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