Question

One mole of ethanol is produced reacting graphite, $H_2$ and $O_2$ together. The standard enthalpy of formation is $-277.7kJmo{l}^{-1}$. Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved

• A. - 277.7
• B. - 555.4
• C. - 138.85
• D. - 69.42
• E. - 1110.8

Answer 1

Answer: (B)

The related equation for the formation of ethanol is.

2 C(graphite)$+3H_2(g)+\frac{1}{2}{O}_2(g)⟶C_2{H}_{5}OH(l)$

$\Delta H_f=-277.7kJ/mol$

When 2 moles of graphite is involved the standard enthalpy of reaction is $-277.7kJ/mol$ $\therefore$ When 4 moles of graphite is involved, standard enthalpy of reaction is $2\times (-277.7kJ/mol)$

$=(-)555.4kJ/mol$