What is the value of Delta H^{circ} for the f | vedclass

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(C) The target reaction is the hydration of ethene: $C_2H_{4_{(g)}} + H_2O_{(l)} \longrightarrow C_2H_5OH_{(l)}$.We can obtain this by subtracting equ

Vedclass · Accepted Answer

$-42.0$

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(C) The target reaction is the hydration of ethene: $C_2H_{4_{(g)}} + H_2O_{(l)} \longrightarrow C_2H_5OH_{(l)}$.We can obtain this by subtracting equation $(i)$ from equation $(ii)$:$(ii) - (i): (C_2H_{4_{(g)}} + 3O_{2_{(g)}}) - (C_2H_5OH_{(l)} + 3O_{2_{(g)}})$ $\longrightarrow (2CO_{2_{(g)}} + 2H_2O_{(l)}) - (2CO_{2_{(g)}} + 3H_2O_{(l)})$This simplifies to: $C_2H_{4_{(g)}} - C_2H_5OH_{(l)} \longrightarrow -H_2O_{(l)}$,which rearranges to $C_2H_{4_{(g)}} + H_2O_{(l)} \longrightarrow C_2H_5OH_{(l)}$.The enthalpy change is $\Delta H^{\circ} = \Delta H^{\circ}_{(ii)} - \Delta H^{\circ}_{(i)} = -1410 \ kJ - (-1368 \ kJ) = -42 \ kJ$.

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Given:
$I$. $2Fe_{(s)} + \frac{3}{2} O_{2(g)} \to Fe_2O_{3(s)}$; $\Delta H^{\Theta} = -193.4 \ kJ$
$II$. $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$; $\Delta H^{\Theta} = -140.2 \ kJ$
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