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For the gas phase reaction,
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
Which of the following conditions are correct ?
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
Which of the following conditions are correct ?
A
$\Delta H \lt 0$ and $\Delta S \lt 0$
B
$\Delta H \gt 0$ and $\Delta S \lt 0$
C
$\Delta H = 0$ and $\Delta S \lt 0$
D
$\Delta H \gt 0$ and $\Delta S \gt 0$
Detailed Solution
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
The dissociation of $PCl_5$ involves the breaking of two P-Cl bonds, which needs energy, so the reaction is endothermic.
$\Delta H = \Delta E + \Delta n_g RT$, where $\Delta n_g$ = change in the number of moles of gaseous products and reactants.
Here $\Delta n_g = 2 - 1 = +1$, i.e. positive, so $\Delta H$ = +ve.
$\Delta S = S_{products} - S_{reactants}$
One mole of gas gives two moles of gas, so the randomness increases and $\Delta S$ = +ve.
Hence $\Delta H \gt 0$ and $\Delta S \gt 0$.
The dissociation of $PCl_5$ involves the breaking of two P-Cl bonds, which needs energy, so the reaction is endothermic.
$\Delta H = \Delta E + \Delta n_g RT$, where $\Delta n_g$ = change in the number of moles of gaseous products and reactants.
Here $\Delta n_g = 2 - 1 = +1$, i.e. positive, so $\Delta H$ = +ve.
$\Delta S = S_{products} - S_{reactants}$
One mole of gas gives two moles of gas, so the randomness increases and $\Delta S$ = +ve.
Hence $\Delta H \gt 0$ and $\Delta S \gt 0$.