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For the reaction : $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$ at constant temperature, $\Delta H - \Delta E$ is :
A
+RT
B
–3RT
C
+3RT
D
–RT
Detailed Solution
$\Delta H = \Delta E + \Delta n_g RT$, so $\Delta H - \Delta E = \Delta n_g RT$, where $\Delta n_g$ = moles of gaseous products $-$ moles of gaseous reactants.
Gaseous products: 3 mol $CO_2$ (water is a liquid, so it is not counted).
Gaseous reactants: 1 mol $C_3H_8$ + 5 mol $O_2$ = 6 mol
$\Delta n_g = 3 - 6 = -3$
$\Delta H - \Delta E = -3RT$
Gaseous products: 3 mol $CO_2$ (water is a liquid, so it is not counted).
Gaseous reactants: 1 mol $C_3H_8$ + 5 mol $O_2$ = 6 mol
$\Delta n_g = 3 - 6 = -3$
$\Delta H - \Delta E = -3RT$