Question

methyl ether, a useful organic solvent, is prepared in two steps. In the first step, carbon dioxide and hydrogen react to form methanol and water:(g) (g) (l) (l)In the second step, methanol reacts to form dimethyl ether and water:(l) (g) (l)Calculate the net change in enthalpy for the format

Answer 1

This question is incomplete, the complete question is;

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) ΔH₁ = -131.kJ

In the second step, methanol reacts to form dimethyl ether and water:

2CH₃OH(l) → CH₃OCH₃(g) + H₂O(l) ΔH₂ = 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.

Answer:

the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ

Step-by-step explanation:

Given the data in the question;

For the First Step;

CO₂(g) + 3H₂(g) → CH₃OH(l) + H₂O(l) ΔH₁ = -131.kJ

For the First Step;

2CH₃OH(l) → CH₃OCH₃(g) + H₂O(l) ΔH₂ = 8.kJ

Using Hess's Law of Constant Heat Summation;

" regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes "

we multiply step 1 reaction by the coefficient of 2

2CO₂(g) + 2×3H₂(g) → 2CH₃OH(l) + 2H₂O(l) ΔH₁ = 2 × -131.kJ

we have

2CO₂(g) + 6H₂(g) → 2CH₃OH(l) + 2H₂O(l) ΔH₁ = -262 kJ

2CH₃OH(l) → CH₃OCH₃(g) + H₂O(l) ΔH₂ = 8 kJ

{ 2CH₃OH cancels 2CH₃OH }

Hence, we have;

2CO₂ + 6H₂ → CH₃OCH₃(g) + 3H₂O(l)

So According to Hess's Law;

ΔH
`_{sum` = ΔH₁ + ΔH₂

we substitute

ΔH
`_{sum` = -262 kJ + 8 kJ

ΔH
`_{sum` = -254 kJ

Therefore, the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ