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:
co₂(g) + 3h₂(g) → ch₃oh(l) + h₂o(l)
δh = -131 kj
in the second step, methanol reacts to form dimethyl ether and water:
2ch₃oh(l) → ch₃och₃(g) + h₂o(l)
δh = 8 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.
kj

Explanation:

Step1: Analyze the reactions

We have two reactions:

1. \(CO_2(g)+3H_2(g)\to CH_3OH(l)+H_2O(l)\), \(\Delta H = - 131\space kJ\)
2. \(2CH_3OH(l)\to CH_3OCH_3(g)+H_2O(l)\), \(\Delta H=8\space kJ\)

We want to find the enthalpy change for the formation of \(1\space mol\) of \(CH_3OCH_3\) from \(CO_2\) and \(H_2\).

Step2: Manipulate the reactions

Multiply the first reaction by \(2\) to get \(2CO_2(g)+6H_2(g)\to 2CH_3OH(l)+2H_2O(l)\), \(\Delta H=-2\times131=-262\space kJ\)

Then add the second reaction \(2CH_3OH(l)\to CH_3OCH_3(g)+H_2O(l)\), \(\Delta H = 8\space kJ\)

Step3: Calculate the net enthalpy change

Using Hess's law, add the enthalpy changes of the two manipulated reactions.
\(\Delta H_{net}=-262 + 8=-254\space kJ\)

Answer:

\(-254\)