Student A pushed on the cart with a force of 500 N. Student B pushed on the other side of the cart. The cart did not move. How m
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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
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 = ΔH₁ + ΔH₂
we substitute
ΔH = -262 kJ + 8 kJ
ΔH = -254 kJ
Therefore, the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ