Question

dimethyl ether, a useful organic solvent, is prepared in two steps. in the first step, carbon dioxide and hydrogen react to form methanol and water: \\(\ce{co_{2}(g) + 3h_{2}(g) \
ightarrow ch_{3}oh(l) + h_{2}o(l)} \quad \delta h = -131.\\,\text{kj}\\) in the second step, methanol reacts to form dimethyl ether and water: \\(\ce{2ch_{3}oh(l) \
ightarrow ch_{3}och_{3}(g) + h_{2}o(l)} \quad \delta h = 8.\\,\text{kj}\\) calculate the net change in enthalpy for the formation of one mole of dimethyl ether from carbon dioxide and hydrogen from these reactions. round your answer to the nearest kj.

Explanation:

Step1: Multiply first reaction by 2

$2CO_2(g) + 6H_2(g)
ightarrow 2CH_3OH(l) + 2H_2O(l)$
$\Delta H_1 = 2 \times (-131\ \text{kJ}) = -262\ \text{kJ}$

Step2: Add second reaction

$2CH_3OH(l)
ightarrow CH_3OCH_3(l) + H_2O(l)$
$\Delta H_2 = 8\ \text{kJ}$
Total: $2CO_2(g) + 6H_2(g)
ightarrow CH_3OCH_3(l) + 3H_2O(l)$
$\Delta H_{\text{total}} = -262 + 8 = -254\ \text{kJ}$

Step3: Check moles of dimethyl ether

The total reaction produces 1 mole of dimethyl ether, so the net enthalpy is $\Delta H_{\text{total}}$.

Answer:

-254