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The enthalpy and entropy change for the reaction:
$Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g)$
are 30 kJ mol$^{-1}$ and 105 J K$^{-1}$ mol$^{-1}$ respectively. The temperature at which the reaction will be in equilibrium is :-
$Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g)$
are 30 kJ mol$^{-1}$ and 105 J K$^{-1}$ mol$^{-1}$ respectively. The temperature at which the reaction will be in equilibrium is :-
A
285.7 K
B
273 K
C
450 K
D
300 K
Detailed Solution
For the reaction $Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g)$: $\Delta H = 30$ kJ mol$^{-1}$ and $\Delta S = 105$ J K$^{-1}$ mol$^{-1}$
At equilibrium, $\Delta G = 0$
$\Delta G = \Delta H - T\Delta S = 0$
$\therefore \Delta H = T\Delta S$
$T = \dfrac{\Delta H}{\Delta S} = \dfrac{30 \times 1000}{105}$ (converting kJ to J)
$T = 285.7$ K
At equilibrium, $\Delta G = 0$
$\Delta G = \Delta H - T\Delta S = 0$
$\therefore \Delta H = T\Delta S$
$T = \dfrac{\Delta H}{\Delta S} = \dfrac{30 \times 1000}{105}$ (converting kJ to J)
$T = 285.7$ K