The combustion enthalpies of carbon,hydrogen, | vedclass

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(C) The combustion reactions are:$(i)$ $C(s) + O_2(g) 	o CO_2(g), \Delta H_1 = -395.5 \, kJ \, mol^{-1}$(ii) $H_2(g) + \frac{1}{2} O_2(g) 	o H_2O(l)

Vedclass · Accepted Answer

$-74.7$

Vedclass · Answer

(C) The combustion reactions are:$(i)$ $C(s) + O_2(g) 	o CO_2(g), \Delta H_1 = -395.5 \, kJ \, mol^{-1}$(ii) $H_2(g) + \frac{1}{2} O_2(g) 	o H_2O(l), \Delta H_2 = -285.8 \, kJ \, mol^{-1}$(iii) $CH_4(g) + 2O_2(g) 	o CO_2(g) + 2H_2O(l), \Delta H_3 = -890.4 \, kJ \, mol^{-1}$We need the enthalpy of formation of methane: $C(s) + 2H_2(g) 	o CH_4(g), \Delta H_f = ?$Using Hess's Law: $\Delta H_f = \Delta H_1 + 2(\Delta H_2) - \Delta H_3$$\Delta H_f = -395.5 + 2(-285.8) - (-890.4)$$\Delta H_f = -395.5 - 571.6 + 890.4$$\Delta H_f = -76.7 \, kJ \, mol^{-1}$.Note: The closest option provided is $-74.7 \, kJ \, mol^{-1}$.

The combustion enthalpies of carbon,hydrogen,and methane are $-395.5 \, kJ \, mol^{-1}$,$-285.8 \, kJ \, mol^{-1}$,and $-890.4 \, kJ \, mol^{-1}$ respectively at $25 \, ^oC$. The value of standard formation enthalpy of methane at that temperature is ..... $kJ \, mol^{-1}$.

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