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) =ΔH8.kJ

Calculate the net change in enthalpy for the formation of one mole of dimethyl ether from carbon dioxide and hydrogen from these reactions.

According to the Law of Hess, the enthalpy of a reaction is the same whether if it takes place in one step or in several steps. This means that if we add a number of reactions to get a final reaction, its enthalpy will be equal to the sum of the individual enthalpies. Also, if we multiply one of the reactions by a factor, we also have to multiply its enthalpy by the same factor

Let's multiply the first step by 2 and add the step 2 to it.

2 CO₂(g) + 6 H₂(g) → 2 CH₃OH(l) + 2 H₂O(l) ΔH= −262 kJ

+

2 CH₃OH(l) → CH₃OCH₃(g) + H₂O(l) ΔH = 8kJ

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2 CO₂(g) + 6 H₂(g) + 2 CH₃OH(l) → 2 CH₃OH(l) + 2 H₂O(l) + CH₃OCH₃(g) + H₂O(l)

2 CO₂(g) + 6 H₂(g) → CH₃OCH₃(g) + 3 H₂O(l) ΔH= -262 kJ + 8 kJ = -254 kJ