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)
`\rightarrow` CH₃OH (l) + H₂O(l); ΔH = -131.kJ In the second step, methanol reacts to form dimethyl ether and water: 2CH₃OH (l)
`\rightarrow` CH₃OCH₃(g) + H₂O (l); ΔH = 8.kJCalculate 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.

Answer 1

ΔH = - 254 kJ

Step-by-step explanation:

The final equation for the enthalpy of formation we want to calculate is:

2 CO₂ (g) + 6 H₂(g) -------------- CH₃OCH₃ (l) ΔH = ?

Notice that if we take the 2 times the first reaction and add the second , by Hess's law we will add the enthalpy changes to obtain our desired ΔH .

( CO₂(g) + 3H₂(g) CH₃OH (l) + H₂O(l); ΔH = -131.kJ) x 2

2CO₂(g) + 6H₂(g) --------- 2CH₃OH (l) + 2 H₂O(l); ΔH = -262.kJ

2CH₃OH (l) --------- CH₃OCH₃(g) + H₂O (l); ΔH = 8.kJ

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